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):
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.
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
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