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### 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

- Doctor Edwin, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
High School Physics/Chemistry

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