Question

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?

Answer 1

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.