Question

The height in feet of a golf ball hit into the air is given by `h= -16t^2 + 64t`, where t is the number of seconds elapsed since the ball was hit. For how many seconds is the ball more than 48 feet up in the air?

Answer 1

Ball is above 48 feet when `t in (1,3)` so for as near as makes no difference ball will spend 2 seconds above 48feet.

Explanation:

We have an expression for `h(t)` so we set up an inequality:

`48 < -16t^2 + 64t`

Subtract 48 from both sides:

`0 < -16t^2 + 64t - 48`

Divide both sides by 16:

`0 < -t^2 + 4t - 3`

This is a quadratic function and as such will have 2 roots, ie times where the function is equal to zero. This means that the time spent above zero, ie the time above `48ft` will be the time in between the roots, so we solve:

`-t^2+4t-3 = 0`

`(-t +1)(t-3) = 0`

For left hand side to be equal to zero, one of the terms in brackets must equal zero, so:

`-t+1 = 0 or t - 3=0`

`t = 1 or t = 3`

We conclude that the golf ball is above 48 feet if `1 < t < 3`