Question

A ball is thrown into the air with an initial upward velocity of 48 ft/s. Its height (h) in feet after t seconds is given by the function \large h=-16t^2+48t+64. After how many seconds will the ball hit the ground? 4 seconds7 seconds5 seconds6 seconds

Answer 1

Given the trajectory of a ball thrown into the air:

$h(t)=-16t^2+48t+64$

To know the time the ball lasts in the air until it hits the ground:

$0=-16t^2+48t+64$

Factoring:

$\begin{gathered} 0=-16(t+1)(t-4) \\ t=-1 \\ t=4 \end{gathered}$

Time can't be negative, so the only solution for the equation is t = 4.

ANSWER

4 seconds