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Solving a Projectile given only the initial height, maximum height, & horiz distanceby Matthew_S
Tags: physics, projectile launcher, projectile motion, projectile problem, projectile velocity 
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#1
Aug1912, 11:56 AM

P: 6

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 


#2
Aug1912, 02:16 PM

P: 172

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 ycomponent of initial velocity (I called initial velocity v_{i}, and the ycomponent is v_{iy}), and set v_{iy} to 0. Solve this for t; I called it t_{max}, 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 t_{max} for t. The height can be called h_{max}, this is one of your known parameters. Solve this equation for v_{i}. Then, you want to make an equation for the final xdistance (I called this x_{f}). The projectile will reach this final xdistance 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 (t_{f}). Make an equation for the xdisplacement using the xcomponent of velocity (v_{ix}=v_{i}cos(θ)). Replacing t with t_{f} and x with x_{f}, this allows you to use another known parameter. Substitute in the v_{i} and t_{f} found earlier, and you can solve for theta. After you have theta, you can plug it into your v_{i} equation to find the numerical value for v_{i}. Hope this helps. 


#3
Aug1912, 04:00 PM

P: 6

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.



#4
Aug1912, 04:35 PM

P: 172

Solving a Projectile given only the initial height, maximum height, & horiz distance
In the height equation, you use v_{iy}=v_{i}sinθ to solve for v_{i}. 


#5
Aug1912, 05:02 PM

Thanks
P: 5,801

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



#6
Aug1912, 05:02 PM

P: 6

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. 


#7
Aug1912, 05:09 PM

Thanks
P: 5,801

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



#8
Aug1912, 05:23 PM

P: 172




#9
Aug1912, 05:23 PM

Thanks
P: 5,801




#10
Aug1912, 05:36 PM

P: 6

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



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