The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = –16t2 – 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?
256 ft/sec

–96 ft/sec

0 ft/sec

112 ft/sec

Question

The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = –16t2 – 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?
 256 ft/sec

–96 ft/sec

0 ft/sec

112 ft/sec

amistre64 · Answer

what are your thoughts about how to find the solution?

anonymous · Answer

I think I am suppose to find the dirivitive.  So -32t-24. But  I am unsure of what to do after that.

akashdeepdeb · Answer

You are right about the derivative. $$v(t) = -32t-64$$
But, we have to find velocity at time 't'. 
Hence, find 't'.

You can find it by setting $s(t) = -16t^2-64t + 80 = 0$, because when the ball hits the ground, s(t) = 0.