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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

V^2 = Vo^2 + 2g*h h = (V^2-Vo^2)/2g h = ((190)^2-(30)^2)/64 = 550 Fi.

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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