# 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?

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 $1 < t < 3$