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

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tykoyz

Answered 2022-07-26
Author has **17** answers

$y=40t-16{t}^{2}$ 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

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

Violet Woodward

Answered 2022-07-27
Author has **2** answers

$y=40t-16{t}^{2}$

dy/dt = 40 - 32t, maximum when $dy/dt=0:40-32t=0\Rightarrow t=1.25s$

maximum height when $t=1.25s\Rightarrow y=40\ast 1.25-16\ast {1.25}^{2}=50-25=25ft$

dy/dt = 40 - 32t, maximum when $dy/dt=0:40-32t=0\Rightarrow t=1.25s$

maximum height when $t=1.25s\Rightarrow y=40\ast 1.25-16\ast {1.25}^{2}=50-25=25ft$

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