SOLUTION: An object is launched straight up into the air 20 feet above the ground. a ball is thrown up into the air an initial velocity of 64 ft/s. find the two times when the ball's heigh

Question 1109886: An object is launched straight up into the air 20 feet above the ground. a ball is thrown up into the air an initial velocity of 64 ft/s.
find the two times when the ball's height is 42 feet above the ground. Does the ball reach a heght of 96 feet? and the time when the ball hits the ground
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The equation is y=-16t^2+64t+20
The maximum is when t=-b/2a or -64/-32 or 2 seconds
when t=2, y=-64+128+20=84, so no, the ball never reaches 96 feet.
For 42 feet above the ground, 42=-16t^2+64t+20
-16t^2+64t-22=0
t=-64+/- sqrt (4096-1408) divided by -32
t=(-1/32)(-64+/- 51.85)
t=-12.15/-32, or 0.38 sec
t=-115.85/-32, or 3.6 sec
when it hits the ground
-16t^2+64t+20=0
t=(-1/32)(-64+/- sqrt (4096+1280); sqrt term is 73.32
t=-137.32/-32, only positive root, or 4.29 seconds.

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