A ball is thrown vertically from the roof of a 64 foot tall building with a velocity of 16 ft/sec, its h
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A ball is thrown vertically from the roof of a 64 foot tall building with a velocity of 16 ft/sec, its h

[From: ] [author: ] [Date: 12-10-13] [Hit: ]
5)^2 = 67.Now,t = 2.So velocity when it reaches the ground, at 2.562 seconds,......
If a ball is thrown vertically upward from the roof of a 64 foot tall building with a velocity of 16 ft/sec, its height in feet after t seconds is s(t)=64+16t-16t^2. What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height0)?

Please explain how you get the answers.

-
v = ds/dt = 16 - 32t

Max height occurs when the velocity = 0 (the ball has stopped moving).

0 = 16 - 32t
t = 0.5 seconds

s(0.5) = 64 + 15(0.5) - 16(0.5)^2 = 67.5m < - - - MAX HEIGHT

Now, it will reach the ground when height = 0
0 = 64 + 16t - 16t^2
t = 2.562 seconds (ignore the negative solution of course)

So velocity when it reaches the ground, at 2.562 seconds, is
v(2.562) = 16 - 32(2.562) = -65.984m/s or 65.984m/s down
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