If a ball is throwndirectly upward with a velocity of 40ft/s, its height (in feet)after t seconds is given by y=40t-16t^2. What is the maximum heightattained by the ball?

Tamara Bryan 2022-07-25 Answered
If a ball is throwndirectly upward with a velocity of 40ft/s, its height (in feet)after t seconds is given by y = 40 t 16 t 2 . What is the maximum heightattained by the ball?
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Answers (2)

tykoyz
Answered 2022-07-26 Author has 17 answers
y = 40 t 16 t 2 take the derivative
y' = 40 -32t set y' = 0
0 = 40 -32t minus 40 to both side
-40 =-32t divide -32 to both side
5/4 =t plug in t in the oringal function
y ( 5 / 4 ) = 40 ( 5 / 4 ) 16 ( 5 / 4 ) 2 simplify
y(5/4) = 50 - 25 = 25
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Violet Woodward
Answered 2022-07-27 Author has 2 answers
y = 40 t 16 t 2
dy/dt = 40 - 32t, maximum when d y / d t = 0 : 40 32 t = 0 t = 1.25 s
maximum height when t = 1.25 s y = 40 1.25 16 1.25 2 = 50 25 = 25 f t
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