A ball is thrown vertically upwards at $5\text{ m/s}$ from a roof top of $100\text{ m}$. The ball B is thrown down from the same point $2\text{ s}$ later at $20\text{ m/s}$. Where and when will they meet?

Well at first I separated the fact that while ball B falls at say $t$ seconds, ball A's time is given by $t+2$ (due to it being thrown 2 seconds prior to ball B).

Given what's given above (the velocity and height)...I wrote down the following equations to find TIME, where lower a and b denote ball A and B:

$$\begin{align}s_a &= s_{0a} + u_a(t+2) + \frac{1}{2}a(t+2)^2 \\ s_b &= s_{0b} + u_bt + \frac{1}{2}at^2\end{align}$$

$S_a(0) = 100$ m, $S_b(0) = 100$ m, $u_a = 5$ m/s, $u_b = 20$ m/s, $a=-9,8$ m/s$^2$

Now plugging those in gives: $S_a = 100 + 5(t+2) + \frac12(-9.8)(t+2)^2$ and $S_b = 100 + 20t + \frac12(-9.8)t^2$

but i tried equalizing the above two equations with all the proper variables replaced by the values provided in the problem, but unfortunately i don't get the right answer.eventually knowing time i can answer to the "where" part of the question.

I'm lost, I'm not even sure if I tackled the problem correctly.

  • $\begingroup$ Hi Desmond, and welcome to Physics Stack Exchange! We're going to need a bit more information to help you out here; in particular, what values do you plug in for each variable, what answer do you get, and what is the expected answer? $\endgroup$ – David Z Jan 1 '12 at 1:35
  • $\begingroup$ Soa = 100m, Sob = 100m, ua = 5m/s, ub = 20m/s, a=-9,8m/s^2 Now plugging those in gives: Sa = 100 + 5(t+2) + 1/2(-9.8)(t+2)^2 and Sb = 100 + 20t + 1/2(-9.8)t^2 -> I edited the post above to reflect this update.. $\endgroup$ – Desmond Jan 1 '12 at 1:43

Well, one of the velocities is negative: the second ball is thrown down, so $u_b = -20$m/s

  • $\begingroup$ You're right. After changing the sign, I tried to do it over again, but failed to get the right answer. Maybe the whole "set-up" is wrong...the correct answer is t=3,78s $\endgroup$ – Desmond Jan 1 '12 at 2:36
  • $\begingroup$ try graphing your two solutions, draw a picture, see if that suggests anything to you. $\endgroup$ – joseph f. johnson Jan 1 '12 at 2:58

OK, I got it. After plugging the initial velocity of ball B which is -20 m/s, i find t=1,78s. However this is the amount of time taken for ball B to reach the same height as ball A. for Ball A, it is t+2 = 1,78+2 = 3,78s (because ball A is thrown 2 seconds prior to ball B).

Given time, i find they join each other at the height of 48,9m.

thank you to joseph f. johnson for pointing out the negative velocity for B.


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