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?

Question

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?

tykoyz · Accepted Answer

y  =  40  t  −  16      t          2       take the derivativey&#039; = 40 -32t set y&#039; = 00 = 40 -32t minus 40 to both side-40 =-32t divide -32 to both side5/4 =t plug in t in the oringal function  y  (  5      /    4  )  =  40  (  5      /    4  )  −  16  (  5      /    4      )          2       simplifyy(5/4) = 50 - 25 = 25

Violet Woodward · Accepted Answer

y  =  40  t  −  16      t          2      dy/dt = 40 - 32t, maximum when   d  y      /    d  t  =  0  :  40  −  32  t  =  0  ⇒  t  =  1.25  smaximum height when   t  =  1.25  s  ⇒  y  =  40  ∗  1.25  −  16  ∗      1.25          2        =  50  −  25  =  25  f  t

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?

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