[tex]\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}[/tex]
t = 0 seconds, when the ball first took off, and t = 2, 2 seconds later.
Answer: [tex]t_1=0\\t_2=2[/tex]
Step-by-step explanation:
We know that the function [tex]h(t)=-16t^2+32t+48[/tex] models the height "h" of the ball above the ground as a function of time "t".
Then, to find the times in which the ball will be 48 feet above the ground, we need to substitute [tex]h=48[/tex] into the function and solve fot "t":
[tex]48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t[/tex]
Factorizing, we get:
[tex]0=-16t(t-2)\\t_1=0\\t_2=2[/tex]
