Question 388044
 how many seconds will the ball be 75 ft haigh after hit 2 feet off the ground and initial upward velocity of 90 ft / sec. 
75h = 2 ho + 90 t - 16 t squared
-----------------
h(t) = -16t^2 + 90t + 2
At h = 75:
-16t^2 + 90t + 2 = 75
-16t^2 + 90t - 73 = 0
*[invoke solve_quadratic_equation -16,90,-73]
----------------------
The ball passes 75 feet twice, ascending at t = 0.982839853961944, and descending at t = 4.64216014603806 seconds.
-------------------
The ball is exactly 75 feet high for zero time, or dt seconds.
It is at or above 75 feet for the difference between the 2 times above,
=~ 3.659 seconds