SOLUTION: chandra is tossing a softball into the air with an underhand motion. the distance of the ball above her hand at any time is given by the function h(t)=32t-16t^2 for 0 < or equal to

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Question 1174046: chandra is tossing a softball into the air with an underhand motion. the distance of the ball above her hand at any time is given by the function h(t)=32t-16t^2 for 0 < or equal to t < or equal to 2. where h (t) is the height and t is the time (in seconds). find the times at which the ball is in her hand and the maximum height of the ball.
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi,
chandra is tossing a softball into the air with an underhand motion. 
the distance of the ball above her hand at any time is given by the function 

h(t)=32t-16t^2 for 0.  h(t) = 0 

    32t - 16t^2 = 0   (back down to the height of your hand)

    32t = 16t^2

            t = 2 sec   time it takes to get back down to the height of your hand

Wish You the Best in your Studies.


Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.

The height function is the parabola


     h(t) = -16t^ + 32t.


It has zeros, when  h(t) = 0,   or


    -16t^2 + 32t = 0.


Factor the expression


     -16t*(t-2) = 0.


The roots are  t= 0  and  t= 2.    (They are time moments, when the ball is at the ground level).


The maximum is reached at the time moment, which is exactly half-way between the roots: t = 1 second.    ANSWER


The height is maximum at this time


    h(t=1) = -16*1^2 + 32*1 = -16 + 32 = 16  feet.    ANSWER

Solved.

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