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

Answer:

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#