Question

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 where t is time (in seconds).

Answer 1

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

$t = 5.0625$

Step-by-step explanation:

Given

$h(t)=-16t^2+80t+5$

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

$0=-16t^2+80t+5$

Reorder

$-16t^2+80t+5 = 0$

Using quadratic formula, we have:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = -16$      $b =80$      $c = 5$

So, we have:

$t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}$

$t = \frac{-80\±\sqrt{6400 +320}}{-32}$

$t = \frac{-80\±\sqrt{6720}}{-32}$

$t = \frac{-80\±82.0}{-32}$

This gives:

$t = \frac{-80+82.0}{-32}$ or $t = \frac{-80-82.0}{-32}$

$t = \frac{2}{-32}$ or $t = \frac{-162}{-32}$

$t = -\frac{2}{32}$ or $t = \frac{162}{32}$

But time can not be negative.

So, we have:

$t = \frac{162}{32}$

$t = 5.0625$

Hence, time to hit the ground is 5.0625 seconds