Projectile Motion- Fixed initial velocity show the particle cant move beyond a curve

**kevinlightman** · December 29th 2009, 04:00 AM

A boy throws a ball at a constant speed V for every throw, show that no matter what angle alpha to the horizontal this boy throws the ball at, the ball cannot move beyond a certain curve in the xy plane. Find the equation of this curve. Model the ball as a particle and forget about air resistance.

This problem has been annoying me for a while now, I believe I have correctly derived all the equations of motion I can, I have the horizontal distance and the vertical distance, I eliminated time from both of these to obtain the path of the ball for a certain angle. I also have the maximum horizontal and vertical distance obtained for a certain angle. I thought to find the maximum distance from the boy that the ball will obtain I should square the horizontal and vertical distances, sum them, differentiate them with respect to t, solve for t for 0 and then substitute t back in, but that didn't get me very far.
Any ideas?

Edit: Apparently the answer is $y=(V^3 - g^2x^2)/(2gV)$ but I can't see how to get there

**Grandad** · December 29th 2009, 07:06 AM

Hello kevinlightman

Originally Posted by kevinlightman

A boy throws a ball at a constant speed V for every throw, show that no matter what angle alpha to the horizontal this boy throws the ball at, the ball cannot move beyond a certain curve in the xy plane. Find the equation of this curve. Model the ball as a particle and forget about air resistance.

This problem has been annoying me for a while now, I believe I have correctly derived all the equations of motion I can, I have the horizontal distance and the vertical distance, I eliminated time from both of these to obtain the path of the ball for a certain angle. I also have the maximum horizontal and vertical distance obtained for a certain angle. I thought to find the maximum distance from the boy that the ball will obtain I should square the horizontal and vertical distances, sum them, differentiate them with respect to t, solve for t for 0 and then substitute t back in, but that didn't get me very far.
Any ideas?

Edit: Apparently the answer is $y=(V^3 - g^2x^2)/(2gV)$ but I can't see how to get there

I have a solution, but it doesn't quite agree with the one you've been given. In fact, I don't see how that solution can be correct. If we check the dimensions:

$V$ is $LT^{-1}$ (i.e. speed is length / time)

$g$ is $LT^{-2}$

$x$ is $L$

So:

$V^3$ is $L^3T^{-3}$

and

$g^2x^2$ is $L^2T^{-4}L^2=L^4T^{-4}$

So the expression $V^3-g^2x^2$ is dimensionally incorrect.

Here's my solution.

Eliminating $t$ between the equations

$x = V\cos\alpha t$

and

$y = V\sin\alpha t - \tfrac12gt^2$

we get

$y = x\tan\alpha - \tfrac12 g \frac{x^2}{V^2}\sec^2\alpha$

$\Rightarrow 2V^2y =2V^2x\tan\alpha-gx^2(1+\tan^2\alpha)$ , using $\sec^2\alpha = 1+\tan^2\alpha$