0

Two people are throwing a ball from shoulder height $S$ in a hallway with ceiling $H$ meters high. The ball is thrown with velocity $v$. Show that the maximum separation is $d=4\sqrt{(H-S)v^2/2g - (H-S)^2}$ if $v^2 \ge 4g(H-S)$ and $d = v^2/g$ if $v^2 \le 4g(H-S)$.

I am able to prove both these results but I have no understanding as to what is actually happening with these results.

I used mamimum height is $H-S = \frac{v^2\sin^2{a}}{2g}$ metres, and range $d=\frac{v^2}{g}\sin{2a}$.

For the $d = v^2/g$ case, it "works" if you just let $a = 45$ degrees. I don't know how the condition $v^2 \le 4g(H-S)$. comes into play though. Same with the other case. I just substitute in $H-S = \frac{v^2\sin^2{a}}{2g}$ to prove $d=4\sqrt{(H-S)v^2/2g - (H-S)^2}$ if $v^2 \ge 4g(H-S)$ but what is the difference between these cases?

user71207
  • 1,543

1 Answers1

1

Note that the second case can be expressed as $H\geq v^2/(4g)+S$, i.e., it's the case of the ceiling being sufficiently high. So I'd interpret this as: If the ceiling is high enough, then the maximum distance is just the usual maximum range. But if the ceiling is too low, then you'll need to angle the shot downwards: this will reduce the range, but it'll also lower the peak height.

Some more details: A quick calculation shows that the maximum height achieved is $y_{max}(\theta)=S+\dfrac{v^2}{2g}\sin(\theta)$, while the range (since the ball start and ends at the same height) is $R(\theta)=\dfrac{v^2}{2g}\sin(2\theta)$. Hence we're looking to maximize $R(\theta)$, subject to the constraint $y_{max}(\theta)\leq H$. Thinking about the behavior of $\sin(2\theta)$ and $\sin(\theta)$ on the range of angles $\theta\in (0,\pi/2)$, we obtain the story I sketched above.

Semiclassical
  • 15,842
  • God, thanks I have been thinking about this for days. Can't believe I did't even think of this. – user71207 Aug 02 '21 at 06:59
  • How does subbing in the height into the distance give you the modified maximum distance though? I get that you will need to change the angle, thus it will have to vary. I still don't get why this is guaranteed by this substitution though $\sin^2{a} = 2g(H-S)/v^2$ – user71207 Aug 02 '21 at 07:05
  • Without looking into the details, I would presume the calculation amounts to find the path whose max height is $H$ above the ground, and then compute the range of this path. (An angle lower than this will reach the ground faster, while an angle higher will pass the ceiling.) If the ceiling is high enough, then no path will reach $H$. Hence in that case you pick the path which maximizes the range. – Semiclassical Aug 02 '21 at 09:01