SOLUTION: Assume that a ball is thrown vertically upward with initial velocity fo 96 ft per second. The distance s(t) (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2.

Question 40038: Assume that a ball is thrown vertically upward with initial velocity fo 96 ft per second. The distance s(t) (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2.
a. at what instant will it be back at the ground level?
b. for what time interval is the ball more than 112ft above the ground?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE WHICH IS SIMILAR TO YOURS AND TRY.
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1. Assume that a ball is thrown vertically upward with initial velocity of
> 96 ft. per second. The distance s(t) (in feet) of the ball from the ground
> after t seconds is s(t) = 96t-16t^2.
>
> a. at what instant will it be back at the ground level?
AT GROUND S=0=96T-16T^2=16T(T-9)
HENCE T=0 ..THAT IS BEFORE IT IS THROWN UP AND
T=9...SO AFTER 9 SECS.IT WILL HIT THE GROUND ON FALL
> b. for what time interval is the ball more than 128 ft. above the ground.
> S=128=96T-16T^2.DIVIDING BY 16
T^2-6T+8=0=T^2-4T-2T+8=T(T-4)-2(T-4)=(T-4)(T-2)=0
T=2 SECS ON THE UPWARD TRAVEL AND
T=4 SECS ON THE DOWNWARD TRAVEL.HENCE IT WILL BE OVER 128 FT.HIGH
DURING T=2 TO 4 SECS.
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