Question 88855
An object that is falling or vertically projected into the air has its height, in feet, above the ground given by s=-16t^2+vot=so Where s is the height, in fee, vo is the original velocity of the object in feet per second, t is the time the object is in motion, is seconds, and so is the height in feet, from which the object is dropped or projected. The figure shows that a ball is thrown straight up from a rooftop at an original velocity of 80 feet per second from a height of 96 feet. The ball misses the rooftop on its way down and eventually strikes the ground. Use the formula and the infromation to solve problem 89-91 I am solving even #90 
s=-16t^2+80t+96 
t=4 
s= -16(4^2)+80(4)+96 
s= -16(16)+320+96 
s=-256+320+96 
s= 160ft

CAN SOMEONE PLEASE HELP ME SOLVE THE SECOND PART OF THIS PROBLEM USING THE EQUATION s=-16t^2+80t+96 How many seconds the ball will take to reach the ground?
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s = 0 when the ball hits the ground.
-16t^2+90t+96 = 0
-2(8t^2 - 45t - 48)=0
(8t^2 - 45t - 48)=0
t = [45 +-sqrt(45^2-4*8*-48)]/16
t = [45 +- 59.67411]/16
t = 6.54 seconds
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Cheers,
Stan H.