Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Free Fall


Date: 10/16/97 at 09:01:22
From: Randy De Souza
Subject: Free Fall

A ball with a mass of 0.5kg is dropped from a height of 190cm.  
The ball is in flight for 5 sec, with gravity acting on it.  
We can assume that the ball will hit the ground and bounce back to 
half the initial height. Given this information determine:

a) How long will it take for the ball to come to rest?
b) How fast will the ball be traveling on the intial drop?
c) Why does the ball change change colour from red to blue?

How do you do this problem?


Date: 10/16/97 at 15:13:18
From: Doctor Rob
Subject: Re: Free Fall

The acceleration of gravity is constant, regardless of the mass.
Call it g cm/sec^2.  Then the acceleration is g, the velocity is
v = g*t + v0, and the height is s = g*t^2/2 + v0*t + s0. Up until
the first bounce, since the ball was "dropped," we assume that
v0 = 0, and we are given that s0 = 190 cm.  If it hits the ground
after t1 = 5 seconds, we have an equation  0 = g*5^2/2 + 190, which
allows us to determine g = -380/25 = -76/5 = -15.2.  Obviously
this is on a small planet, since at the surface of Earth,
g = -980 cm/sec^2.

Between the first and second bounce, we have s0 = 0, so
s = -7.6*(t-5)^2 + v0*(t-5), so s = 0 when t = t1 or t1 + v0/7.6,
and the maximum height will occur halfway in between, at
t = 5 + v0/15.2, so 95 = -7.6*(v0/15.2)^2 + v0^2/15.2, which
allows us to solve for v0 = Sqrt[95*30.4] = 38*Sqrt[2] = 53.74,
so the time of the second bounce is t = t1 + 5*Sqrt[2] = 12.07.
Let t2 = 5*Sqrt[2], the time taken for the second bounce.

Between the second and third bounce, a similar analysis tells us
that the time is t3 = 5. Then t4 = 5/Sqrt[2]. In fact one can
show that t(n+1)/t(n) = 1/Sqrt[2] for all n. The time until the
ball comes to rest is T = 5 + Sum t(n), where the sum is over all
n > 1. Then the sum is a geometric progression, so you can add it
up, getting T = 5 + 5*Sqrt[2]/(1 - 1/Sqrt[2]) = 15 + 10*Sqrt[2],
or about 29.142 seconds.

Maximum absolute speed on the initial drop is 15.2*5 = 76 cm/sec.

The ball must turn color because of the heating caused by the
conversion of potential and kinetic energy into heat, which happens
because the ball doesn't rebound to its full height from which it was
dropped.  The initial energy of the ball is all potential, which you
can calculate.  Why red and blue, I have no idea.

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

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/