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?

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
$y=40t-16{t}^{2}$ 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
$y\left(5/4\right)=40\left(5/4\right)-16\left(5/4{\right)}^{2}$ simplify
y(5/4) = 50 - 25 = 25
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Violet Woodward
$y=40t-16{t}^{2}$
dy/dt = 40 - 32t, maximum when $dy/dt=0:40-32t=0⇒t=1.25s$
maximum height when $t=1.25s⇒y=40\ast 1.25-16\ast {1.25}^{2}=50-25=25ft$