The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 48 feet high is s (t) = - 16t2 - 32t + 48, 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?

Remember that w hen the ball hits the ground, s (t) = 0

So we have: - 16t^2 - 32t + 48 = 0

It's time to determine t = ( - (-32) + / - sqrt ((-32) ^2 - 4 (-16) (48))) / (2 (-16))

t = (32 + / - sqrt (1024 + 3072)) / (-32)

t = (32 + / - sqrt (4096)) / (-32)

t = (32 + / - 64) / (-32)

t = (48 or - 32) / (-32)

t = - 1.5 or 1

Since a negative number doesn't make sense, so t = 1

Keep in mind that velocity is the derivative of the height, therefore:

s' (t) = - 32t - 32

s' (t) = - 32*1 - 32

s' (t) = - 32 - 32

s' (t) = - 64

So the answer 64 feet per second dwn