# Solving Projectile w/ Initial Height, Height Max, Horiz Dist.

• Matthew_S
In summary, designing a formula for determining the launch velocity, launch angle, and time spent in the air for a projectile can be challenging when the initial launch height and ground level are different. One approach is to use a system of equations to find the initial velocity and launch angle, and then use these to calculate the horizontal displacement. Another approach is to use the potential/kinetic energy equation and solve for the vertical and horizontal components of velocity separately. However, in some cases, there may not be a solution for the projectile's flight time and distance traveled.
Matthew_S
Hi everyone,

I'm trying to design a formula that determines the launch velocity, launch angle, and time spent in the air for a projectile if only the initial launch height, maximum height reached, and total horizontal distance traveled before the projectile hits the ground are known. It's not too hard to design a formula if the initial height is the ground (since the parabolic path taken by the projectile is symmetrical), but once the initial height and ground level are different, things get a lot harder. I tried looking everywhere online and deriving my own formula, without success. Does anyone know of such a formula?

Thanks,
Matt

You can work through it with a system of equations but it gets a bit convoluted. I'll explain the process that you'd go through to solve it, but I won't bother writing out the equations as they get quite long and ugly. Let me know if further explanation is needed.

The ball reaches the max height when the vertical velocity is 0, so make an equation for the y-component of initial velocity (I called initial velocity vi, and the y-component is viy), and set viy to 0. Solve this for t; I called it tmax, for the time that the projectile reaches the max height.

Next, make an equation for the height of the projectile as a function of time. In this equation, substitute tmax for t. The height can be called hmax, this is one of your known parameters. Solve this equation for vi.

Then, you want to make an equation for the final x-distance (I called this xf). The projectile will reach this final x-distance at the moment the height of the projectile reaches 0, and it hits the ground. Using your height equation and setting height to 0, you can solve for t, to find the time that the ball contacts the ground (tf).

Make an equation for the x-displacement using the x-component of velocity (vix=vicos(θ)). Replacing t with tf and x with xf, this allows you to use another known parameter. Substitute in the vi and tf found earlier, and you can solve for theta.

After you have theta, you can plug it into your vi equation to find the numerical value for vi. Hope this helps.

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Thank you so much! I don't think you solve for the velocity in the height equation, only the vertical component. But using your steps, I was able to get the angle to cancel out and get only the velocity. The ugliest kinematics equation I've ever seen, but it gets the job done.

Matthew_S said:
Thank you so much! I don't think you solve for the velocity in the height equation, only the vertical component. But using your steps, I was able to get the angle to cancel out and get only the velocity. The ugliest kinematics equation I've ever seen, but it gets the job done.

If you cancel out the angle, it won't do you any good because your answer is dependent on the angle.
In the height equation, you use viy=visinθ to solve for vi.

From the launch height h and max height H, using the potential/kinetic energy equation, you get $Y = \sqrt {2g(H - h)}$, where $Y$ is the vertical component of the velocity. You also get $T_1 = \frac Y g = \sqrt { \frac { 2(H - h) } g }$ for the time of ascent, and, similarly, $T_2 = \sqrt { \frac { 2H } g }$ for the time of descent. Then, $X = \frac D {T_1 + T_2}$, for the horizontal velocity component, and $\theta = \arctan \frac Y X$. You could probably simplify things somewhat.

When the angle canceled out, I was left with an equation whose only unknowns were the total time and velocity. I managed to express t in terms of h_max, h_0, and fundamental constants and solved for v, which could then be used for theta. In response to voko, I didn't see that approach until now but it looks a lot more efficient than what I did before.

I'm now working on solving a projectile problem knowing only the initial launch height, the maximum launch height, and the time the projectile was in the air. I'm trying to find the distance by working backwards, but it's not working so far.

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For the sake of completeness: $V = \sqrt {X^2 + Y^2}$ - that's the magnitude of the launch velocity.

Matthew_S said:
When the angle canceled out, I was left with an equation whose only unknowns were the total time and velocity. I managed to express t in terms of h_max, h_0, and fundamental constants and solved for v, which could then be used for theta. In response to voko, I didn't see that approach until now but it looks a lot more efficient than what I did before.

The angle can cancel out of the max height equation because with the right velocity, you can reach a given max height from any launch angle. However, the xf location is different for each launch angle in this case. There's only one angle that will make a trajectory that goes through both points (hmax and xf), so although it may seem like the angle can cancel from the height equation, it's necessary to keep it in.

Matthew_S said:
I'm now working on solving a projectile problem knowing only the initial launch height, the maximum launch height, and the time the projectile was in the air. I'm trying to find the distance by working backwards, but it's not working so far.

There is no solution for this. The total flight time is a condition only on the vertical velocity component. The horizontal component may be any, and you can get arbitrary distance for any flight time.

That's true, thanks again - you saved me a lot of time

## 1. What is the formula for solving projectile motion with initial height, maximum height, and horizontal distance?

The formula for solving projectile motion with initial height, maximum height, and horizontal distance is y = y0 + (ym - y0)(x/xm) - (x2/xm2)(g/2), where y is the vertical position, y0 is the initial height, ym is the maximum height, x is the horizontal distance, xm is the horizontal distance at maximum height, and g is the acceleration due to gravity.

## 2. How do you determine the initial velocity in a projectile motion problem with given initial height, maximum height, and horizontal distance?

To determine the initial velocity, you can use the formula v0 = √(g(xm2 + 2(ym - y0))), where v0 is the initial velocity, g is the acceleration due to gravity, xm is the horizontal distance at maximum height, ym is the maximum height, and y0 is the initial height.

## 3. How can you find the time of flight for a projectile with known initial height, maximum height, and horizontal distance?

The time of flight can be found using the formula t = xm/v0, where t is the time of flight, xm is the horizontal distance at maximum height, and v0 is the initial velocity.

## 4. Can you use the same formula for solving projectile motion with initial height, maximum height, and horizontal distance in both horizontal and vertical directions?

No, the formula for solving projectile motion with initial height, maximum height, and horizontal distance can only be used for the vertical direction. For the horizontal direction, the formula x = x0 + v0t can be used, where x is the horizontal position, x0 is the initial horizontal position, v0 is the initial horizontal velocity, and t is the time.

## 5. What is the significance of knowing the maximum height in a projectile motion problem with given initial height, maximum height, and horizontal distance?

The maximum height is important because it tells you the highest point the projectile reaches in its trajectory. It can also be used to determine the initial velocity and time of flight for the projectile. Additionally, knowing the maximum height can help in predicting the range of the projectile and its landing point.

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