# Homework Help: Difference quotient problem and average rate of change

1. Jul 15, 2013

### needingtoknow

1. The problem statement, all variables and given/known data

A soccer ball is kicked into the air from a platform. The height of the ball, in metres, t seconds after it is kicked is modelled by h(t) = -4.9t^2 + 13.5t + 1.2.

I solved for the expression that represents the average rate of change of height over the interval 2<= t <= (2+h) and got which is correct (according to the answer key):

[-4.9(2+h)^2 + 13.5(2+h) - 7.4] / h

For what values of h is the expression above not valid?

The answer in the back states h>0.84; at 2.84 s, h = 0.

I understood how they got h>0.84 by finding the zeroes of the quadratic equation to get 2.84 and then subbing that into the inequality to get the change (h) to be 0.84. Therefore since t !> 2.84 h !> 0.84. But how do they get at 2.84 s, h != 0 as restrictions?

How did they get 0.84 and 2.84 in the restrictions. This question is a bit confusing to me. Can anyone help me find the answer?

Last edited: Jul 15, 2013
2. Jul 15, 2013

### Staff: Mentor

It is confusing that h is both used as the height of the object and a time.

h(t)=0 (the height) corresponds to the ball landing on the floor. This will certainly stop its free fall, and make the formula for h(t) invalid after that point.