Question 1065245
.
Using physics, it can be shown that if a ball is thrown upwards with an initial velocity
of 16 ft/s from the top of a building 64 ft high, then its height h above the ground t
seconds later will be
h(t) = 64 + 16t − 16t2
During what time interval will the ball be at least 32 ft above the ground?

I got [0,2] but I am confused on the actual time interval. 
~~~~~~~~~~~~~~~~~~


{{{graph( 330, 330, -1.5, 5.5, -10.5, 80.5,
          -16x^2 + 16x +64, 32
)}}}


Plots h(t) = {{{-16t^2 + 16t +64, 32}}} (red), h = 32 (green)


Is it more clear now ?