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Radda [10]
3 years ago
11

A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. Its height​ h, in​ feet, after t seconds is

given by ​h(t)=−16t^2 +96t +640. After how long will the ball reach the​ ground?

Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0
Check the picture below.

thus is at 0 = -16t² + 96t +640,

\bf 0=-16t^2+96t+640\implies 0=-16(t^2-6t-40)
\\\\\\
0=t^2-6t-40\implies 0=(t-10)(t+4)\implies t=
\begin{cases}
\boxed{10}\\
-4
\end{cases}

well, clearly it can't be a negative value for the elapsed seconds, so it can't be -4.

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Step-by-step explanation:

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Step-by-step explanation:

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