Question

Former NFL punter Ray Guy holds the record for the longest hangtime on a punt. If the ball leaves with an upward velocity of 128 ft/s from an initial height of 4 feet, how long will the ball be in the air? Use the formula where h is the height of the ball in feet and t is the time in seconds since it is kicked. Round your answer to the nearest tenth.

Answer 1

Answer:

The correct option is D.

Step-by-step explanation:

The given equation

$y=-16t^2+128t+4$

Where,  h is the height of the ball in feet and t is the time in seconds since it is kicked.

The leading coefficient is negative, so it is a downward parabola.

First of all find zeros of given function.

$0=-16t^2+128t+4$

Using quadratic formula we get

$t=\frac{-128\pm \sqrt{(128)^2-4(-16)(4)}}{2(-16)}$       $[\because t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}]$

$t=8.031,-0.031$

So the ball is in the air between t=-0.031 and t=8.031. Since the time can not be negative, therefore the ball is in the air between t=0 and t=8.031.

The ball is in the air for 8.031 second, so option D is correct.