Aball is thrown into the air with an upward velocity of 20 feet per second it's height h in feet after t seconds is given by the function h(t)=16t^2+20t+2. how long does it take the ball to reach its maximum height? what is the ball's maximum height?

The extreme of the quadratic ax² +bx +c is found at x = -b/(2a). Your function has
  a = 16
  b = 20
so it will have an extreme value at t = -20/(2*16) = -20/32 = -5/8.

We assume you intend h(t) = -16t² +20t +2 (with a leading minus sign), in which case the extreme value (maximum height) occurs at t = 5/8. The maximum height is
  h(5/8) = (-16*5/8 +20)(5/8) +2 = 50/8 +2 = 33/4 = 8.25

The ball reaches its maximum height after 5/8 = 0.625 seconds.
The ball's maximum height is 8 1/4 = 8.25 feet.

The answer is: a = about 63.59ft2 for 15

A=bxh3.2x1.8=5.76 centimeters square