The function  H(t) = -16t^2 + 64t + 80  is a quadratic function, whose plot is a parabola opened down.


This quadratic function has the maximum at the value of its argument  t = , where "a" is the coefficient at t^2

and "b" is the coefficient at t.


In your case, a= -16,  b= 64, so the function gets the maximum at  t =  = 2.


So, the ball gets the maximum height 2 seconds after is is thrown straight up. 


To calculate the maximum height, substitute t= 2 seconds into the given formula

    H(t) = -16*2^2 + 64*2 + 80 = 144 feet.


Now the last question: when the ball hits the ground.


For it, you need to equate H(t) to zero, which is the ground level

    H(t) = 0,  or  -16*t^2 + 64t + 80 =0.


Simplify the equation

    16t^2 - 64t - 80 = 0

cancel the common factor 16

    t^2 - 4t - 5 = 0

and factor the left side

    (t-5)*(t+1) = 0.


The only positive root t= 5 is the solution.  The ball will hit the ground in 5 seconds.