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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?

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antonsandiego
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

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