Question 24750
a) s(t) = -16t^2 + 64t 


b) s(1) = -16(1)^2 + 64(1) = 48 feet


c) s(t) = 0 at ground level, so 
-16t^2 + 64 t= 0


Solve by factoring the common factor of -16t:
-16t ( t - 4) = 0
t= 0 or t= 4


Time to reach the ground is from t= 0 to t=4 which is 4 seconds.


Maximum height will be halfway between t=0 and t=4, which is t= 2 seconds to reach the maximum height.  Maximum height will be 
s(t) = -16t^2 + 64 t
s(2) = -16(2)^2 + 64(2)
s(2) = -64 + 128 = 64 feet


R^2 at SCC