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GrogVix
3 months ago
15

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48feet above the ground,

the function h(t)=−16t2+32t+48 models the height, h, of the ball above the ground as a function of time, t. Find the times the ball will be 48feet above the ground.
Mathematics
2 answers:
babunello [11.8K]3 months ago
5 0

\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}

At t = 0 seconds, the ball is released, and at t = 2 seconds, it reaches the same height again.

PIT_PIT [12.4K]3 months ago
5 0

Answer: t_1=0\\t_2=2

Step-by-step explanation:

We’re given the function h(t)=-16t^2+32t+48 that describes the ball’s height "h" above ground over time "t".

To determine when the ball is at 48 feet, substitute h=48 into the equation and solve for "t":

48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t

After factoring, we find:

0=-16t(t-2)\\t_1=0\\t_2=2

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