SOLUTION: A ball is thrown vertically upward from the top of a building 96feet tall with an initial velocity of 80feet per second. The distance S (in feet )of the ball from the ground, after
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Question 460356: A ball is thrown vertically upward from the top of a building 96feet tall with an initial velocity of 80feet per second. The distance S (in feet )of the ball from the ground, after t seconds is s=96+80t-16tsquard.
(a)After how many seconds does the ball strike the ground?
(b) After how many seconds will the ball pass the top of the building on its way down? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically upward from the top of a building 96feet tall with an initial velocity of 80feet per second. The distance S (in feet )of the ball from the ground, after t seconds is s=96+80t-16tsquard.
(a)After how many seconds does the ball strike the ground?
When s = 0 it hits the ground
s=96+80t-16t^2 = 0
-t^2 + 5t + 6 = 0
(-t + 6)*(t + 1) = 0
t = -1 second (Ignore)
t = 6 seconds to impact
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(b) After how many seconds will the ball pass the top of the building on its way down?
96+80t-16t^2 = 96
-t^2 + 5t = 0
t = 0 (at launch)
t = 5 seconds (on the way down)
