Gravitational Catch-Up

Date: 09/15/2003 at 14:34:05
From: Paavana
Subject: Motion of a particle under gravity

I was doing a question from my homework, which I've been puzzling 
over with no success! The question is: 

  A stone is dropped from the top of a tower. One second 
  later another stone is thrown vertically downwards from 
  the same point with a velocity of 14 m/s. If they hit 
  the ground together find the height of the tower.

I'm so confused! I would like to know how to go about answering 
questions of this nature in case I should encounter one again.

Date: 09/16/2003 at 12:32:16
From: Doctor Edwin
Subject: Re: Motion of a particle under gravity

Hi, Paavana.

This is the kind of question that can be very difficult to solve 
unless you think about it in just the right way. I know this because 
I spent about a day thinking about it until I found the right 
approach.

Once both stones are dropped, their relative velocity doesn't change. 
That is, they're both speeding up, but at the same 9.8m/sec^2. Let's 
call the time you dropped the first stone t_1, and the time you threw 
the second one t_2, and t_3 will be the time they hit the ground 
(which is also the time that the second stone catches up with the 
first).

So if you know the speed difference between the first stone and the 
second, and how far ahead the first stone was at t_2, then you can 
figure out how long it took the second stone to catch up with the 
first (that's t_3 - t_2).

We're given the speed of the second stone at t_2. Can you tell how 
fast the first stone was moving at t_2? How long would it take the 
second stone to catch up with the first? How long was the first stone 
falling? How far would it have fallen in that time?

Is that clear? Please write back if you're having difficulty and I'll 
be happy to help you.

- Doctor Edwin, The Math Forum
  http://mathforum.org/dr.math/