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Sveta_85 [38]
3 years ago
8

A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds af

ter being launched is s(t) = −16t 2 + 32t+ 240 feet above the ground. What is the velocity of the ball as it hits the ground?
Physics
1 answer:
Vilka [71]3 years ago
8 0

Answer:

V = (-32t +32) ft/s

The velocity of the bowling ball after t seconds is the time derivative of the position function s(t). To obtain the velocity from s(t) we simply differentiate s(t) with respect to t. That is

V = ds(t)/dt = -16×2t + 32×1 = -32t + 32.

When differentiating, you multiply the coefficient of each term in the equation with the power of the variable and then reduce the power by 1 just like above.

Explanation:

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Hope this answers your question.

5 0
4 years ago
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Answer:

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First Quarter Moon is the second primary Moon phase and it is defined as the moment the Moon has reached the first quarter of its orbit around Earth, hence the name. It is also called Half Moon as we can see exactly 50% of the Moon's surface illuminated. Whether you see the left or right half illuminated, depends on several factors, including your location.

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@timeanddate.com

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Explanation:

4 0
3 years ago
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Answer:

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