The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = â16t2 â 64t + 80, where t is the time elapsed that the ball is in the air. what is the instantaneous velocity of the ball when it hits the ground?

The height of the ball (ft), measured upward from the ground is
s(t) = 16t² + 64t + 80
where t = time when the ball is in the air.

When the ball hits the ground, s = 0. The time(s) when this occurs is given by
16t² + 64t + 80 = 0
Divide through by 16.
t² + 4t + 5 = 0
(t - 1)(t + 5) = 0
t = 1 or t = -5 s
Reject negative time, so that
t = 1 s

The velocity function is
v(t) = 64 + 32t
When t = 1 s, obtain
v(1) = 96 ft/s

The ball hits the ground with velocity 96 ft/s.

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