An object is dropped from the top of ABC Tower, 1,500 feet from the ground.

An object is dropped from the top of ABC Tower, 1,500 feet from the ground.  Calculate the average velocity of the object over the interval t=2 seconds and t=3 seconds.  The position function of the projectile is s(t) = -16t2 + V0t + h0, where V0t is the initial velocity of the object and h0 is the height to which it is dropped.
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1 Answer

Putting h0=1500 and v0=0, height after t=2=-16*4+1500=1500-64=1436'.

Height after t=3=-16*9+1500=1500-144=1356'.

Distance travelled is 1436-1356=80' in 1 second. So the average velocity (downwards) is 80'/s.

by Top Rated User (1.1m points)

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