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

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

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 $\displaystyle y=(V^3 - g^2x^2)/(2gV)$ but I can't see how to get there

2. 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 $\displaystyle y=(V^3 - g^2x^2)/(2gV)$ 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:
$\displaystyle V$ is $\displaystyle LT^{-1}$ (i.e. speed is length / time)

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

$\displaystyle x$ is $\displaystyle L$
So:
$\displaystyle V^3$ is $\displaystyle L^3T^{-3}$
and
$\displaystyle g^2x^2$ is $\displaystyle L^2T^{-4}L^2=L^4T^{-4}$
So the expression $\displaystyle V^3-g^2x^2$ is dimensionally incorrect.

Here's my solution.

Eliminating $\displaystyle t$ between the equations
$\displaystyle x = V\cos\alpha t$
and
$\displaystyle y = V\sin\alpha t - \tfrac12gt^2$
we get
$\displaystyle y = x\tan\alpha - \tfrac12 g \frac{x^2}{V^2}\sec^2\alpha$

$\displaystyle \Rightarrow 2V^2y =2V^2x\tan\alpha-gx^2(1+\tan^2\alpha)$, using $\displaystyle \sec^2\alpha = 1+\tan^2\alpha$
Re-arrange as a quadratic in $\displaystyle \tan\alpha$:
$\displaystyle gx^2\tan^2\alpha - 2V^2x\tan\alpha +2V^2y +gx^2=0$
Now the values of $\displaystyle \tan\alpha$ that satisfy this quadratic must be real. So:
$\displaystyle (-2V^2x)^2-4gx^2(2V^2y+gx^2)\ge0$

$\displaystyle \Rightarrow 4V^4x^2 \ge 4gx^2(2V^2y+gx^2)$
Divide by $\displaystyle 4x^2$, which we may do since $\displaystyle x^2\ge 0$:
$\displaystyle \Rightarrow V^4 \ge 2gV^2y + g^2x^2$

$\displaystyle \Rightarrow y \le \frac{V^4-g^2x^2}{2gV^2}$
Thus the values of $\displaystyle x$ and $\displaystyle y$ are bounded by the curve
$\displaystyle \Rightarrow y = \frac{V^4-g^2x^2}{2gV^2}$
which, I think, is the correct answer. (Unless you can see where I've made a slip?)