Question 143954
You are given a equation that relates distance (s) to both time (t) and initial velocity (V0)

{{{s = -16t^2 + v[0]t}}}

The first question asks you to find the equation for time when s = 80 for a given v0.

{{{80 = -16t^2 + v[0]t}}}
{{{0 = -16t^2 + v[0]t - 80}}}
Now use the quadratic equation with a=-16, b=vo and c = -80
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

Note that you won't get 'an answer' that is a number since the value v0 is not known. If it turns out that there is only one time when s=80, then you know that 80 feet up is as high as the projectile goes. If there are two times, then one is on the way up and the other is on the way back down.  You won't know yet since you have not been given a value for V0. But you can find what v0 will result in s=80 being the highest point (just find when {{{sqrt( b^2-4*a*c )}}} is 0

Either way, I suggest you move from where the projectile was fired from, because it is going to return there :) When will it return there? Well that is problem 2.

When will it hit the ground?
Use the same logic as in problem 1, except instead of 80 feet, use 0. There will be two answers for 0. one will be at time 0, the other is the one they want.