Time Interval of Ball in the Air?

Question

A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 96t - 16t^2. For what time interval is the ball more than 112 feet above the ground?

Answer 1

2* sqrt2=2*1.414=2.828 seconds.

One finds the 2 solutions to the quadratic equation

      112= 96t -16t^2, or in the form at^2 +bt +c =0

t^2 -6t +7 =0

Then use the general solution:

t= {-b +/-sqrt(b^2-4ac)}/2a

two solns are

t2=3+sqrt(2)

t1=3-sqrt(2)

The time in the air above 112 feet is t2-t1=2*sqrt(2)