* The classical example of problem solving, maximizing the range of a projectile launched from height h with velocity v over the ground level, has received various solutions*. In some of these, one can find the maximization of the range R by differentiating R as a function of an independent variable or through the implicit differentiation in Cartesian or polar coordinates The classical example of problem solving, maximizing the range of a projectile launched from height h with velocity v over the ground level, has received various solutions If a projectile is launched at a speed u from a height H above the horizontal axis, and air resistance is ignored, the maximum range of the projectile is Rmax = u g√u2 + 2gH, where g is the acceleration due to gravity. The angle of projection to achieve Rmax is θ = arctan(u √u2 + 2gH). Can someone help me derive Rmax as given above The path of this projectile launched from a height y 0 has a range d.. In physics, assuming a flat Earth with a uniform gravity field, and no air resistance, a projectile launched with specific initial conditions will have a predictable range.. The following applies for ranges which are small compared to the size of the Earth

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees. Now suppose that instead of a flat surface we launch the projectile off of a cliff as shown below. Also, I am using the typical convention that g = 9.8 N/kg = 9.8 m/s2 so that the acceleration in the y-direction is -g Maximum Height As the projectile travels through air, it climbs up to some maximum height (h) and then begins to come down. At the instant when the projectile is at the maximum height, the vertical component of its velocity is zero. This is the instant when the projectile stops to move upward and does not yet begin to move downward

I agree with the formula given by Mr. Alexander Blake for the range d. However, the statement You basically want to maximize sin(2θ) is not correct. There is a factor (sinθ)^2 inside the square root. To maximize d, you have to calculate d(d)/dθ. * The maximum range for projectile motion*. A projectile of the same mass can be launched with the same initial velocity and different angles \(\theta_0\). Consider the figure given below where a projectile is launched with three different angles \(45^0,\,\,60^0,\,\,and\,\,30^0\)

- How to find the maximum height of a projectile? Maximum height of the object is the highest vertical position along its trajectory. The object is flying upwards before reaching the highest point - and it's falling after that point. It means that at the highest point of projectile motion, the vertical velocity is equal to 0 (Vy = 0)
- Maximum Projectile Height Formula The following formula describes the maximum height of an object in projectile motion. As noted before, this is without air resistance. h = V₀² * sin (α)² / (2 * g
- When the projectile reaches a vertical velocity of zero, this is the maximum height of the projectile and then gravity will take over and accelerate the object downward. The horizontal displacement of the projectile is called the range of the projectile, and depends on the initial velocity of the object
- Physics Ninja looks at the kinematics of projectile motion. I calculate the maximum height and the range of the projectile motion
- The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees. How do you get this
- We launched a projectile from various heights and measured the horizontal range it travelled before reaching the ground. We concluded that as height increases, range also increases. We are now trying to explain why an increase in launch means an increase in range

The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60-degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first. An analysis of the velocity components for these three projectiles reveals reasons for these observations Solving for the initial height of a projectile in Physics Stack Exchange, and; Projectile Motion Range, Initial Height, and Maximum height in YouTube, which all seem to give answer to your problem. EDIT. Also Maximum range of a projectile (launched from an elevation) here, at Math SE, seems to be the same problem as yours The maximum height reached by the object is 47.9 meters The time of flight is the interval between when the projectile is launched (t 1) and when the projectile touches the ground (t 2) We deﬁne the projectile problem as follows: a projectile is launched from a tower of height h, with initial velocity v, and at an angle measured with respect to the horizontal. We aim to ﬁnd m , the launch angle that maximizes horizontal distance

- A projectile is launched from a height of $h$ and the initial velocity of $\\sqrt{2ga}$. Find the maximum range achieved by the projectile in terms of $g$,$a$, and $h.
- Maximum Range of a Projectile Launched from a Height—C.E. Mungan, Spring 2003 reference: TPT 41:132 (March 2003) Find the launch angle q and maximum range R of a projectile launched from height h at speed u. The basic equations of kinematics at the landing point after flight time T are 0 1 2 =+ -hT gT2 uy (1) vertically and RT= ux (2.
- 1 Range of Projectile Motion 1.1 Horizontal Range Most of the basic physics textbooks talk about the horizontal range of the projectile motion. It is derived using the kinematics equations: a x = 0 v x = v 0x x = v 0xt a y = g v y = v 0y gt y = v 0yt 1 2 gt2 where v 0x = v 0 cos v 0y = v 0 sin Suppose a projectile is thrown from the ground.
- Use one-dimensional motion in perpendicular directions to analyze
**projectile**motion. Calculate the**range**, time of flight, and**maximum****height****of****a****projectile**that is**launched**and impacts a flat, horizontal surface. Find the time of flight and impact velocity of a**projectile**that lands at a different**height****from**that of launch - The maximum height h of a projectile launched with initial vertical velocity v 0y is given by [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex]. The maximum horizontal distance traveled by a projectile is called the range
- A cannon sits on top of a plane at a height ##h## above the ground and fires a shell with an initial velocity ##v_0##. At what angle ##\theta_0## must it fire the shell to attain maximum range ##R_{\text{max}}## along the ground? range maximization for a projectile launched at a fixed vertical distance h above the ground where it eventually.
- Maximum Height In Projectile Motion Definition Projectile motion is a 2D motion that takes place under the action of gravity. When any object is thrown from the ground at a certain angle in an upward direction, it follows a particular curved trajectory

- At maximum height the projectile will only have horizontal component that is v x = u cos θ v y 2 − u y 2 = 2 a y v y = 0 ( a t max h e i g h t H ) u y = u sin θ a y = − g H P u t t i n g t h e s e v a l u e s , 0 = ( u s i n θ ) 2 − 2 g H H = 2 g u 2 sin 2
- The angle of projection at which the horizontal range and maximum height of projectile are equal is A projectile is fired from the level ground at angle View solution. A body is launched up an inclined plane of inclination 3 0.
- In this example, the maximum height reached by the projectile and the range of a projectile was calculated for a projectile launched from a height H above the ground and at an angle a o above the horizontal line. Calculate the time of flight, range, R, and maximum height, y MAX, reached by the projectile for the following special cases

The projectile in this application is launched from an elevation, and yet 45 degrees still produces the maximum range for the projectile! From my understanding the maximum range of projectile depends on configuring the launch to maximum amount of hang-time. Is this application flawed - or am I going crazy Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance)

This maximum horizontal distance that a projectile travels is called the range. The time taken by the projectile to reach the ground from the point of projection is called Time of flight. The maximum vertical distance that it can reach is called maximum height Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance) ** Non-horizontally launched projectiles: This is when an object is thrown in the air from the ground, rises to peak height and then drops to the ground again)**. In this case, at the maximum height, the vertical velocity vy is zero. The particle may still be moving horizontally but is not moving up or down Maximum height is given by H = u²sin²θ/2g And Range is given by R = u²sin2θ/g Here, u is initial velocity of projectile , θ is angle between initial velocity and horizontal line and g is acceleration due to gravity. At θ = 45° Maximum height , H = u²sin²45°/2g = u²(1/√2)²/2g H = u²/4g -----(1

Projectile Motion... Using a firing angle of 45 degrees and a muzzle velocity of 100 meters/second the maximum height is 255.1 meters at a distance of 509.1 meters, the maximum distance is 1018.2 meters Calculate the range of the projectile. The range of the projectile is the total horizontal distance traveled during the flight time. Again, if we're launching the object from the ground (initial height = 0), then we can write the formula as R = Vx * t = Vx * 2 * Vy / g. It may be also transformed into the form: R = V² * sin (2α) / I know that if a projectile is launched from level ground to the same height the most efficient angle is 45 degrees, but as the launching height increases the angle for the maximum range decreases. That's why shot putters want to throw the shot at an angle less than 45 degrees. They are throwing the shot slightly above the ground

If we are on a flat surface, the angle of maximum range is 45 ∘. However, as we increase the initial height (h) of the projectile, we need to throw it at a shallower angle in order to get maximum distance (less than 45 ∘). Also on Cupcake Physics Blog ExhibitingConfidences.com is for sale Students love the **range** equation in introductory physics, but it's really kind of silly. Here is a better way to calculate the **maximum** **range** **of** **a** **projectile** The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. That means that the best way to launch a high-altitude projectile is to send it flying at a 90-degree angle to the ground—straight up The projectile will decelerate on its way to maximum height, come to a complete stop at maximum height, then starts its free fall descent towards the ground. If you use the vertical component of its initial speed, you can write v2 h max =0 = v2 0y −2 ⋅ g ⋅ hmax This is equivalent t Moreover, the maximum horizontal range is achieved with a launch angle which is much shallower than the standard result, . Figure 11: Projectile trajectories in the presence of air resistance. Figure 11 shows some example trajectories calculated, from the above model, with the same launch angle, , but with different values of the ratio

** As we can see from the graphs below, the projectile impacts the ground after approximately 0**.6 seconds. It reaches its maximum height after approximately 0.3 seconds. Its range is approximately 2.4 meters. In approximately 0.3 seconds it has covered half its range AIPMT 2012: The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is (A) θ = tan - 1 ( This video illustrates how to compute the angle that maximizes the range for a projectile that is launched from a certain height and ends up below or above that height. (Part 1 Purpose: The text makes some claims regarding the range of a projectile launched at an angle, namely: the maximum range of the projectile occurs at a launch angle of 45 o with respect to the horizontal.; a projectile launched at an angle with respect to the horizontal will have the same range as a projectile launched at an angle of 90 o - .; This experiment will test these predictions

- e: a) the time of flight (5) b) the range of the projectile (5) c) the maximum height above the ground (5) d) the components of vx and vy when the projectile strikes the ground (5
- ed path that is dependent upon the launch velocity, launch angle, height of release, air resistance and gravity (Pyke 1991). Most of these factors cannot be easily changed; the one exception is the launch angle. In a vacuum, the maximum distance is obtained for a launch angle of 45°, an
- ed by both the sight height, or the distance of the line of sight above the bore centerline, and the range at which the sights are zeroed, which in turn deter
- 2 - Projectile Motion Calculator and Solver Given Range, Initial Velocity, and Height Enter the range in meters, the initial velocity V 0 in meters per second and the initial height y 0 in meters as positive real numbers and press Calculate. The outputs are the initial angle needed to produce the range desired, the maximum height, the time of flight, the range and the equation of the path of.
- Often question about projectiles will ask you to find the maximum height, or the range of the projectile, or even how long it took for the projectile to hit the ground. Maximum Height. When the projectile reaches its maximum height its vertical component of velocity will be zero. We can use the linear equations of motion to calculate this.
- ed as follows: For the vertical part of the motion \(u^2_y\) = \(u^2_y\) + 2a y s. Here u y = usinθ, a = -g, s = h max, and at the maximum height v = 0. Time of flight (T f)

Let us consider projectile range further. Figure 5. Trajectories of projectiles on level ground. (a) The greater the initial speed v 0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0 on the range of a projectile with a given initial speed The initial velocity is v 0 feet per second and the path of the projectile is modeled by the parametric equations. and . Use a graphing utility to graph the paths of a projectile launched from ground level at each value of θ and v 0. For each case, use the graph to approximate the maximum height and the range of the projectile Until we get the following for the maximum range of the projectile: € R max = v g (v2+2gH) To find the angle for the maximum range, plug this into the quadratic equation that started this: € tanθ= −R±R2−4 −gR2 2v2 H− gR2 2v2 2 −gR2 2v 2 = − v g v2+2gH ±0 −g v v2 g2 (v2+2gH) Simplifying this gives the following expression. Projectile Motion Derivation: We will discuss how to derive Projectile Motion Equations or formula and find out how the motion path or trajectory looks like a parabola under the influence of both horizontal and vertical components of the projectile velocity. We will also find out how to find out the maximum height, time to reach the maximum height, the total time of flight, horizontal range. Ans: Maximum height reached = 45.92 m, time of flight = 6.12 s, horizontal range = 318.1 m Example - 03: A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same leve

• A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile Calculate the height of each trial as well. Try to hit a target! d = (v 2 / g)sin(2θ) Where d is the distance traveled, in meters. v is the firing velocity, in meters/second. g is the acceleration due to gravity and θ is the measure of the firing angle, measured from the horizontal. You can also turn the equation around to find maximum height. A projectile, in other words, travels the farthest when it is launched at an angle of 45 degrees. But what about trying to maximize a projectile's height to increase hang time The path of this projectile launched from a height y 0 has a range d.. In physics, assuming a flat Earth with a uniform gravity field, and no air resistance, a projectile launched with specific initial conditions will have a predictable range.. Contents. Ideal projectile motion; Derivations; Actual projectile motion; Projectile characteristic Projectile height given time. This is the currently selected item. Deriving max projectile displacement given time. Deriving max projectile displacement given time. Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization..

- Projectile Motion: Final Velocity, Launch Height & Range This is part of the HSC Physics Syllabus Module 5: Advanced Mechanics. Let's look into some examples of questions involving projectile motion
- what is the meaning of projectile motion If a ball launched with initial velocity of 25 m/s at 45 degree. Find the maximum range covered by the ball? Pls solve what is lowest speed of an object in the projectile motion fired with an angle 30º and speed is 5m/s? curve of radius 120 m is banked at an angle of 18°
- I'm trying to get Matlab to return the maximum range and angle for a projectile launched from a set height h and initial velocity vO. Follow 65 views (last 30 days
- Deriving the maximum range and angle of a projectile Answered. In a thread in the Green group, AnarchistAsian and I were discussing his coil guns. I posed the question of what range he could get, and he asked me to go through the physics derivation. In the vertical direction, the maximum height H = v v t - ½gt 2 (you need calculus to.
- If you calculate the range for a projectile launched at 30 o, you will find it's the same as a projectile launched at 60 o. The same goes for 40 o and 50 o. The graph of range vs angle is symmetrical around the 45 o maximum. The equations used to find out various parameters are shown below; Time of flight, Maximum height, Horizontal range
- The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is asked May 16, 2019 in Physics by Ruksar ( 68.8k points
- Horizontal Range and Maximum Height of a Projectile Before embarking on some examples, let us consider a special case of projectile motion that occurs often. Assume a projectile is launched from the origin at t i 5 0 with a positive v yi component as shown in Figure 4.9 and returns to the same hori-zontal level

** The maximum height of a projectile launched with initial vertical velocity is given by The maximum horizontal distance traveled by a projectile is called the range **. The range of a projectile on level ground launched at an angle above the horizontal with initial speed is given b Since the objective of the question is to calculate range, the 1.63 s value is the one used to find range. To find range, the initial horizontal velocity solved earlier (u x) will be used in the following equation: s x = u x t s x = (17.1)(1.63) The range is 27.9 m. Projectiles That are Launched: Maximum Height The path of this projectile launched from a height y has a range d. in physics , assuming a flat earth with a uniform gravity field , and no air resistance , a, the increase in range if the projectile is fired not lated in range d25:d425. the formulas are writ-ten in application for simulating projectile motion Pg. 78, Derivation 3.1: Maximum height and range of a projectile. Preparatory Questions Please discuss with your partners and write the answers to these in your notebooks. 1. To quote Bauer and Westfall (reference above, top of page 79): The range, R, of a projectile is defined as the horizontal distance between the launching point and th A shot is fired with an angle of 45° with the horizontal with a velocity of 300 ft/s. Find the maximum height and range that the projectile can cover, respectively. A. 800 ft, 1600 ft ; B. 923 ft, 3500 ft ; C. 700 ft, 2800 ft ; D. 1800 ft, 3000 ft; Problem Answer: The maximum height and range that the projectile can cover is 700 ft, 2800 ft

A horizontally launched projectile's initial vertical velocity is zero. Solve the following problems with if the height of the table is doubled, range increases by range of 3m high table is What is the ball's maximum height? Conceptually, ball travels up for 1s at average vertical velocity of 5m/s ( Hi, I have been trying to solve a problem,i need to write code to an existing projectile class to calculate the maximum height reached by a projectile ,i have tried all possible solutions but seem stuck,here is the code belo Discussion: The text makes several claims regarding the range of a projectile launched at an angle, namely: the maximum range of the projectile occurs at a launch angle of 45 o with respect to the horizontal.; a projectile launched at an angle q with respect to the horizontal will have the same range as a projectile launched at an angle of 90 o - q

The range, R, is the greatest distance the object travels along the x-axis in the I sector. The initial velocity, v i, is the speed at which said object is launched from the point of origin.The initial angle, θ i, is the angle at which said object is released.The g is the respective gravitational pull on the object within a null-medium. = The height, h, is the greatest parabolic height. For the Range of the Projectile, the formula is R = 2* vx * vy / g; For the Maximum Height, the formula is ymax = vy^2 / (2 * g) When using these equations, keep these points in mind: The vectors vx, vy, and v all form a right triangle. You can express the horizontal distance traveled x = vx * t, where t refers to time Maximum Height: Range: The maximum height is obtained at the point where the vertical component of the velocity vanishes. At this point, the velocity of the projectile is identical in magnitude and direction to the horizontal component of the velocity. The first solutions correspond to the time when the projectile was launched, t = 0. The. Why is the range for a projectile max at 45 degrees? The equation for range is vcos*t= range. As cosine increases 0= 1 30= .9 45= .7 60= .5 90= 0 so as the angle increases the range decreases? but that is wrong as I know but why mathmatically. Princeton review says the range is this formula

- The formula for range R of a projectile is: R = ((Vo^2)*(sin2θ)) / (g) The horizontal displacement D corresponding to the maximum height is simply half of the range, so. D = R / 2. Now all we need is the initial velocity Vo and the angle θ from which it was launched with respect to the horizontal. We are given the following data
- Rajasthan PMT 2006: If maximum height and range of a projectile are same, what is the angle of projection? (A) 30° (B) 76° (C) 50° (D) 90
- A projectile is launched at the same height at which it lands (ho =hf). If the time of the flight is 20 s, at what time will it be at its maximum height? 10s (1/2 the time of the flight

Projectile motion (horizontal trajectory) calculator finds the initial and final velocity, initial and final height, maximum height, horizontal distance, flight duration, time to reach maximum height, and launch and landing angle parameters of projectile motion in physics The maximum height \(\displaystyle h\) of a projectile launched with initial vertical velocity \(\displaystyle v_{0y}\) is given by \(\displaystyle h=\frac{v^2_{0y}}{2g}\). The maximum horizontal distance traveled by a projectile is called the range. The range \(\displaystyle R\) of a projectile on level ground launched at an angle. The maximum range is a result of high horizontal velocity, and long hang time due to the incline which the projectile is launched. To find an angle at which range and height are maximized would be called the optimal angle. This angle exists at 45 degrees. As displayed above, the vertical and horizontal velocities are equally high, and the hang.

(**Maximum** **height** **of** **a** **projectile** motion Lab Report, n.d.) This experiment's main objective is to determine the velocity with which a **projectile** is **launched** with. To show that the **range** is the same for complimentary angles 4. To show that the trajectory is a parabola Discussion of the four aspects of the experiment The first part. The horizontal range of the projectile is seven times the maximum height of the projectile ie, {eq}R\ =\ 7\times H{/eq}. We are asked to calculate the launch angle of the projectile

Summary: Derived equations for a projectile launched from level ground with initial velocity v i at an angle above the ground: Time of flight g v t i f 2 sin Time to height f i h t g v 21 sin Maximum height g v h i 2 sin2 Range g v R i 2 sin2 If the ground is not level, for example throwing a ball from the top of a building, thes The range of a projectile launched with speed v 0 at an angle θ 0 with respect to the horizontal x-axis is d = v 0 2 sin(2θ 0)/g. For two projectile launched with the same speed v 0 to have the same range they have to be launched at angles θ ± = π/4 ± θ. Details of the calculation: Assume rock 1 is thrown from point A with speed v 0 and.

Play this game to review Physics. A ball thrown straight up takes 8 total seconds to complete its journey all the way up and down. 1 second into its journey and 7 seconds into its journey the magnitude of its velocity will be _____ and its direction _____ We can find the time when a projectile reaches its maximum height by setting v y = v y0 - gt = 0 and solving for t. We find t max_height = v y0 /g = v 0 sinθ 0 /g. We can now find the range R by substituting t = 2t max_height into the equation for x(t). R = v 0 cosθ 0 2t max_height = (2v 0 2 cosθ 0 sinθ 0)/g = (v 0 2 sin2θ 0)/g ( a ) The maximum height occurs for vertical launch of a projectile, which corresponds to a projection. launch angle of 90 degrees. ( b ) The maximum range of a projectile can be shown to occur for a projectile launch angle of 45 degrees How to calculate the time of projectile motion. First, determine the maximum height of the object. This is often the height the object is launched from. Next, use the formula to calculate the time of flight. Using the formula above, you can enter the height to determine the time of flight The position of a projectile launched from the origin at t = 0 is given by = mat t=2s. if the projectile was launched at an angle from the horizontal, then is (take g = 10 14927574 2.2k

Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface. \n; Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch. \n; Calculate the trajectory of a projectile. \ From this we need to obtain expressions for the range and the maximum height in order to solve the problem. To find the range, we need to determine the time of flight. This will give us two times, one at t = 0 which corresponds to before the projectile is launched and the other which shall be when it lands - the time of flight! 0 = -1/2 g t. Answer to: A projectile is launched with velocity v = 60.0 m/s at an angle theta = 30 degrees above the horizontal. What is its maximum height.. Solving projectile problems with quadratic equations Example: A projectile is launched from a tower into the air with initial velocity of 48 feet per second. Its height, h, in feet, above the ground is modeled by the function h = -16t 2 + v 0 t + 64 where t is the time, in seconds, since the projectile was launched and v 0 is the initial velocity The maximum height of a projectile is greatest when the launch angle is largest. C. A projectile has maximum range when launched at an intermediate angle, around 45°. D. The range of a projectile is the same whether launched at an angle or at the complement of that angle. (Two angles are complementary if they add up to 90°; for example, 25.

Range of a projectile, including air resistance. Introduction Here we study the motion of a projectile thrown through the air, including the important effects of air resistance.We will investigate how the maximum distance the projectile travels before hitting the ground (optimized with respect t The launch velocity of a projectile can be calculated from the range if the angle of launch is known. It can also be calculated if the maximum height and range are known, because the angle can be determined. From the range relationship, the launch velocity can be calculated Using the Pythagorean Theorem, you can use the sides to figure out the total speed with which the ball was launched. You can use the same triangle to figure out the angle at which it took off. Figuring out the height comes back to just worrying about the ball's vertical motion. We know how quickly the ball left your foot At 0˚ elevation ( firing the projectile horizontally from ground level) the range will be zero because the projectile will fall a tiny distance to hit hit the ground immediately it is fired. At 90˚ elevation ( firing the projectile straight up ) the range will once again be zero as the projectile falls back to starting position max range is at 45o and 2 complimentary launch angles at the same speed gave the same range For a projectile launched over level ground DERIVE EXPRESSIONS for time of flight (t), max height ( y max), and horizontal range ( x) in terms of v 0, , and g. Confirm that the derived equations predict the observations above

2. The range of a projectile is an important consideration in various applications. If the landing point is at the same height as the launch point then the range of the projectile depends upon the initial velocity, the angle of projection, and gravity. 2 sin 2 o v R g θ = Show that the range has a maximum value when θ = 45° The maximum height above the ground of a projectile launched at an angle of 29.2 to the horizontal is found to be 57.5 m. What is the initial speed of the projectile?b. What is the horizontal range of a projectile launched with a speed of 39.6 m/s at an angle of 28.2 above the horizontal?c Problem 23 Hard Difficulty. A projectile is fired with an initial speed of 200 m/s and angle of elevation $ 60^\circ $. Find (a) the range of the projectile, (b) the maximum height reached, and (c) the speed at impact