Posted by Sam on Wednesday, July 24, 2013 at 1:27pm.
A ball is thrown straight downward from the top of a tall building, with an initial speed of 30 ft/s. Suppose the ball strikes the ground with a speed of 190 ft/s. How tall is the building? Acceleration is 32 ft/s squared

calculus  Henry, Wednesday, July 24, 2013 at 2:25pm
V^2 = Vo^2 + 2g*h
h = (V^2Vo^2)/2g
h = ((190)^2(30)^2)/64 = 550 Fi.

calculus  Reiny, Wednesday, July 24, 2013 at 5:05pm
height = 16t^2 30t + d, where d is the height of the building
d(height)/dt = velocity = 32t  30
190 = 32t  30
32t = + 160
t = 5
at t = 5 , height = 0
0 = 16(25)  150 + d
550 = d
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