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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.

y=-16x^2+147x+80

1 Answer

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Answer:

t = 9.7 seconds

Explanation:

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation as :


y=-16x^2+147x+80

We need to find the time that the rocket will hit the ground, to the nearest 100th of second.

When the rocket hit the ground, So,


-16x^2+147x+80=0\\\\x=-0.515\ s\ and\ 9.70\ s

Neglecting negative tim,

t = 9.7 s

Hence, it will hit the ground at 9.7 seconds.

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