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If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 96 ft/sec, then what is the maximum height the ball reaches? what is the velocity of the ball when it hits the ground?\

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The height function would be s(t) = -16t^2 + 64t + 32.

Maximum height ---> set s'(t) = 0.

-32t + 64 = 0. ---> t = 2. s(2) = -16(2)^2 + 64(2) + 32 = the maximum height is 96 feet.

To find the velocity when it hits the ground, set s(t) = 0.

-16t^2 + 64t + 32 = 0. Divide by -16.

t^2 - 4t - 2 = 0. The only positive solution is x = 4.45.

s'(4.45) = -32(4.45) + 64 = It's going -78.4 feet/second when it hits the ground.
User Rohit Dhawan
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