Calculus
posted by Anonymous on .
A ball is thrown upward with an initial velocity of 96 feet per second from the top of a 100 foot building. (a) When well the ball attain its maximum height? (b) When will the ball hit the ground? Give a decimal answers correct to 2 decimal places.

the height relative to the ground will be
s(t) = 16t^2 + 90t + 100
a)
s'(t) = 32t + 90
= 0 for max of s(t)
32t = 90
t = 90/32 = 2.8125 seconds
b)
ball hits the ground when s(t) = 0
16t^2 + 90t + 100=0
8t^2  45t  50 = 0
solve for t using the quadratic formula , refect the negative result 
Shouldn't 90t be 96t since 96 feet/second is the velocity?