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 when 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

Idon’t understand what you’re asking..