Rock Falling From Building

Date: 9/12/96 at 18:16:21
From: Carl C. Smith
Subject: Rock Falling From Building of Height...

Dear Dr.Math,

A rock is dropped from a tall building and hits the ground at 
224 ft/sec (approximately 150 miles/hr).  How long did it take 
for the object to fall?

I tried doing the equation, which is T=D/V.  I can't do that without 
knowing how tall the building is, or without knowing how long it 
takes for the rock to reach terminal velocity.  If you could write 
back and send some tips I would really appreciate it.  
					
Sincerely,
     Amber

Date: 9/12/96 at 18:56:8
From: Doctor Tom
Subject: Re: Rock Falling From Building of Height...

Hi Amber,

The problem is that you are using the wrong formula.  The one you're 
using assumes that the rock is moving at the same speed all the time, 
but for your problem, it starts not moving (the instant before you let 
go), and it winds up moving very fast indeed.

You need two formulas (which you can probably find in your book
somewhere).  One says that the speed at any time is:

V = At,  where V is the speed, A is the acceleration of gravity
(maybe it's called "g"), and t is the time.  You know v when it
hits (224 ft/sec) and you know A (32 ft/sec^2).  So you can work
out t = V/A = 224/32 = 7 seconds.

The other formula you need tells how far a rock will fall, starting
from standing still, under constant acceleration.  That formula is:

D = 1/2(At^2)

D will be the height of the building, A is 32, and t (you just
worked it out) is 7.  So D = 32*49/2 = 784 feet.

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 9/15/96 at 20:52:20
From: Carl C. Smith
Subject: Re: Rock Falling From Building of Height...

Doctor Tom,

Thanx so much for your help, I really appreciate it.

Sincerely,
Amber