SOLUTION: From the top of the building in the illustration, a ball is thrown straight up with an initial velocity of 40 feet per second. The equation s = -16t2 + 40t + 48 gives the height s

Question 537209: From the top of the building in the illustration, a ball is thrown straight up with an initial velocity of 40 feet per second. The equation s = -16t2 + 40t + 48 gives the height s of the ball t seconds after it is thrown. Find the maximum height reached by the ball and the time it takes for the ball to hit the ground. (Round your answer to one decimal place.)
Thank you in advance!
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
From the top of the building in the illustration, a ball is thrown straight up with an initial velocity of 40 feet per second. The equation s = -16t2 + 40t + 48 gives the height s of the ball t seconds after it is thrown. Find the maximum height reached by the ball and the time it takes for the ball to hit the ground. (Round your answer to one decimal place.)
s = -16t2 + 40t + 48
The "vertex" defines the max height.
It reaches this height when
t = -b/(2a)
t = -40/(2(-16))
t = -40/(-32)
t = 1.25 seconds
.
Max height:
s = -16(1.25)^2 + 40(1.25) + 48
s = 73.0 feet
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