# Finding the maximum height a ball can be dropped from and still collide with a ball thrown below it

I don't necessarily need a specific answer, but I could use a hint, direction, or maybe some reading material.

The question states:

A rubber ball is shot straight up from the ground with speed $V(0)$. Simultaneously, a second rubber ball at height $h$ is directly above the first ball is dropped from rest.

a) At what height above the ground do both balls collide. Your answer will be an algebraic expression in terms of $h$, $V(0)$, and $g$.

b) What is the maximum value of $h$ for which a collision occurs before the first ball falls back to the ground?

c) For what value of $h$ doe the collision occur at the instant when the first ball is at its highest point?

I have solved A and found the answer to be: $$d = h - \frac{gh}{2V(0)}$$ EDIT: After redoing the problem, the answer comes out to be: $$d=h-\frac{gh^2}{2V(0)^2}$$ I believe this is correct, but I am completely stumped on questions b and c.

EDIT: I'm also not sure whether or not $h$ is meant to represent the height the second ball starts at or the distance between the balls at any given moment. It seems I haven't solved the first part correctly, so I'll post my work on it:

I plugged the variables I received into the function:

$$Distance = d_0 + V_0 + \frac{1}{2}at^2$$

(where $d_0$ is the starting distance, $V_0$ is the starting velocity, $a$ is acceleration, $t$ is time), to describe the position functions for the two balls as:

$$d_0 = V(0)t - \frac{1}{2}gt^2$$

$$d_1 = h - \frac{1}{2}gt^2$$

I set $d_0 = d_1$, but I'm not sure what I should be solving for. I can solve it as:

$$V(0)t = h$$

But I can't find any kind of use for this.

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## 2 Answers

(a) I assume that you tackled the problem more or less like this.

Let ground level be $0$, and let the "up" direction be positive. If $U(t)$ is the displacement at time $t$ of the ball that was thrown up, then $$U(t)=V_0t -\frac{1}{2}gt^2,$$ until the ball returns to ground. If $D(t)$ is the displacement at time $t$ of the ball that was dropped, then $$D(t)=h-\frac{1}{2}gt^2.$$ Without worrying yet about excessively large values of $t$, where the equations do not apply, we can set $U(t)=D(t)$. After simplifying, we get a very nice equation. solve for $t$ and substitute in either equation to find the collision height.

(b) The first ball returns to the ground at time $t=\dfrac{2V_0}{g}$. The problem asks us for the maximum height $h$ so that the two balls collide before the first hits the ground. There is technically no such maximum height. But we can find the $h$ so that they collide exactly when the first (and second) hits the ground. Just set the $t$ you found in part (a) equal to $\dfrac{2V_0}{g}$.

(c) The ball thrown up reaches is highest point at $t=\dfrac{V_0}{g}$. Apart from that, it is same calculation as in (b).

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I think there's a typo in the $t$-value in (c). –  Arthur Sep 27 '12 at 6:19
@Arthur: Indeed there was. Fixed. Thanks! –  André Nicolas Sep 27 '12 at 6:25
Thank you very much. I believe I'm beginning to understand this. The answers I found using your solutions were: A. d = h - (g*h^2)/(2*V(0)^2) B. h = (2*V(0)^2)/g C. h = (V(0)^2)/g –  FullR Sep 27 '12 at 7:03
Why do you say "there is technically no such maximum height" in part (b)? Surely if the second ball is dropped from very high, the first one can go up and come back down before they collide (on the ground). Besides, the math shows a clear maximum height, as sketched in my answer. –  Jonathan Sep 28 '12 at 18:48
Question (b) says before the ball hits the ground. Maximum height is if the balls collide exactly when they hit the ground. Even if the hitting is a millisecond before ground is hit, height $h$ is not maximum. But if you change the before in (b) to "before or when" then there is a maximum $h$. I am just being picky about wording. A simpler physics question. What is the maximum speed reached by the dropped ball before it hits the ground? No such max speed. It keeps speeding up until it hits. –  André Nicolas Sep 28 '12 at 18:59

First, doublecheck your answer. Dimensionally, it doesn't make sense. $gh/v_0$ doesn't have dimensions of length.

Next, you find that they collide at some height $d$, and the problem requires that $d\ge 0$. This inequality is solved for certain values of $h$.

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I redid my work and found my answer to A should be: d = h - ((g*h) / (2*V(0)^2) I think. –  FullR Sep 27 '12 at 6:01
@JMeyers: It should be $h-\frac{1}{2}g\frac{h^2}{V_0^2}$, jut a slip. –  André Nicolas Sep 27 '12 at 6:12
@AndréNicolas Thank you. I don't know how I managed to make so many little mistakes. –  FullR Sep 27 '12 at 6:13
@JMeyers: I am older, so am confident I have made a lot more little mistakes (and big ones) than you. –  André Nicolas Sep 27 '12 at 6:18