Jason is standing on the ground and throws a ball vertically upward with an initial speed of 80 ft/sec.   its height after t seconds is given by h = 80t-16t^2  a) how high does the ball go?   b)   how many seconds does it take to reach maximum height? c) when does the ball hit the ground?    

Hello, I think you are missing extra information, another formula to go with this problem. 
Using calculus,
the velocity equation will be the derivative of the h function that gives.
v=80-32t

At its highest point the velocity=0, so we set v=0 and solve for t.
0=80-32t
32t=80   t=2.5 seconds
the height at this point would be
h=80(2.5)-16(2.5)^2
h=200-16(6.25)=200-100=100
A) h=100ft

B) at t=2.5 seconds

The ball will take an additional 2.5 seconds to come back down. Therefore the total time for it to hit the ground would be 
C) 5 seconds.

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!