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### Rate of acceleration/gravity

```
Date: 04/09/97 at 16:09:34
From: Catherine Mangan
Subject: Rate of acceleration/gravity

My question is: If a ball is dropped from a building and hits the
ground after 3 seconds, how tall is the building?

I think that the rate of acceleration is 9.8 meters/second or
something like that.

My science teacher gave us the information that at one second, the
ball is traveling 9.8 meters per scond, at two seconds the ball is
traveling at a rate of 19.6 meters/second, and when the ball hits the
ground at three seconds, the terminal velocity was 29.4 meters per
second.

I would appreciate your help on this question, for I really have no
idea how to solve it. Thank you.
```

```
Date: 04/10/97 at 15:45:43
From: Doctor Anthony
Subject: Re: Rate of acceleration/gravity

Hello Catherine,

Neglecting air resistance the ball will experience a constant
acceleration of 9.8 m/sec.  There are four very useful CONSTANT
acceleration formulae which can be used for this type of problem.  Any
textbook on elementary mechanics will show you the derivation of these
formulae, so look them up.

If u = initial velocity
v = final velocity
s = distance
a = acceleration
t = time

The formulae are     s = (1/2)(u+v)t  ..........(1)

v = u + at   ..............(2)

s = ut + (1/2)at^2 ........(3)

v^2 = u^2 + 2as  ..........(4)

In the problem you gave we have   u = 0
a = 9.8
t = 3
s = ?

Knowing u, a, and t, formula (3) will give us the value of s (= height
of building)

s = 0 x 3 + (1/2) x 9.8 x 9

= 44.1 metres.

If you want the terminal velocity use formula (2) v = u + at

= 0 + 9.8 x 3

= 29.4 m/sec

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Physics/Chemistry

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