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RoseWind [281]
3 years ago
14

A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’

s height h at time t seconds is h = -16t2 + 272t + 1344. When will the ball strike the ground?
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
5 0

When the ball is in the ground, its height is basically equal to zero thus making our equation,

<span>    -16t^2 + 272t + 1344 = 0</span>

Simplifying the equation will give us,

<span>  - t^2 + 17t + 84 = 0 or t^2 – 17t – 84 = 0</span>

Factoring out the equation will give us,

<span>   (t – 21)(t + 4) = 0</span>

Thus, t = 21 or t = -4. -4 is an extraneous root. Thus, the answer is t = 21.

<span>Answer: 21 seconds</span>

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\boxed{The\:value\:of\:x\:is\:7.375.}

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

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{ \bf{ \underbrace{To\:verify :}}}

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Hence verifed.

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}

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