Question

WILL GIVE BRAINLIEST!!

A batter hits a baseball upward with an initial speed of 96 feet per second. After how many seconds does the ball hit the ground? Use the formula h = rt − 16t2 where h represents height in feet and r represents the initial speed.

Numerical Answers Expected! Answer for Blank 1:

Answer 1

check the picture below.

since it hits the ground when h(t) = 0, then

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{96}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+96t+0\implies \stackrel{h(t)}{0}=-16t^2+96t\implies 0=-16t(t-6) \\\\\\ \begin{cases} 0=-16t\implies 0=t\\\\ 0=t-6\implies \boxed{6}=t \end{cases}$

two values, since supposedly its initial height is 0, it hits it twice, at the starting point and in the end.