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/