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Susan throws a softball upward into the air at a speed of 32 feet per second from a 104-foot platform. The distance upward that the ball travels is given by the function d(t)=-16t^2+32t+104. What is the maximum height of the softball? How many seconds does it take to reach the ground after first being thrown upward?

When the softball attains maximum height, d(t)/ dt =0

ie -32t +32 =0

t=1 s

When t=1, height = -16 (1)^2 +32 (1) +104 =120 feet (shown)

When it reaches the ground, height above ground = 0 ft

ie -16 t^2 +32t +104 =0

2t^2 -4t - 13 =0 (dividing both sides by -8)

Hence, t= 3.74s (shown) or t=-1.74s (rejected)

Hope this helps. Peace.

Calculus

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