# 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?

##### 1 Answer
Aug 2, 2016

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

#### Explanation:

We have an expression for $h \left(t\right)$ so we set up an inequality:

$48 < - 16 {t}^{2} + 64 t$

Subtract 48 from both sides:

$0 < - 16 {t}^{2} + 64 t - 48$

Divide both sides by 16:

$0 < - {t}^{2} + 4 t - 3$

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 $48 f t$ will be the time in between the roots, so we solve:

$- {t}^{2} + 4 t - 3 = 0$

$\left(- t + 1\right) \left(t - 3\right) = 0$

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

$- t + 1 = 0 \mathmr{and} t - 3 = 0$

$t = 1 \mathmr{and} t = 3$

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