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38t 16t. What is the maximum height attained by the ball? (Round your answer to theIf a ball is thrown directly upward with a velocity of 38 ft/s, its height (in feet) after t seconds is given by ynearest whole number.)ft

Question
38t 16t. What is the maximum height attained by the ball? (Round your answer to the
If a ball is thrown directly upward with a velocity of 38 ft/s, its height (in feet) after t seconds is given by y
nearest whole number.)
ft
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38t 16t. What is the maximum height attained by the ball? (Round your answer to the If a ball is thrown directly upward with a velocity of 38 ft/s, its height (in feet) after t seconds is given by y nearest whole number.) ft

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Step 1

Step 2

At maximum height the velocity is 0 that is the derivative of the height is 0.

d
y38r-162
dt
y38-32
(y0)
38-327 0
32r 38
t 1.1875
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d y38r-162 dt y38-32 (y0) 38-327 0 32r 38 t 1.1875

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Step 3

...
y38(1.1875)-16(1.1875)
= 45.125 -22.5625
-22.5625
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y38(1.1875)-16(1.1875) = 45.125 -22.5625 -22.5625

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Tagged in

Math

Calculus

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