Step 1 :We are given the equation of motion for the projectile as \(s=-16 t^{2}+v_{0} t\). We are asked to find the time(s) when the projectile reaches a height of 64 feet and when it returns to the ground. The initial velocity \(v_{0}\) is given as 80 feet per second.
Step 2 :To find the time(s) when the projectile reaches a height of 64 feet, we need to set \(s=64\) and solve for \(t\).
Step 3 :The times when the projectile reaches a height of 64 feet are 1 second and 4 seconds.
Step 4 :The projectile returns to the ground when \(s=0\). So, we need to set \(s=0\) and solve for \(t\).
Step 5 :The times when the projectile returns to the ground are 0 seconds (which is the initial launch time) and 5 seconds.
Step 6 :Final Answer: (a) The times when the projectile reaches a height of 64 feet are \(\boxed{1, 4}\) seconds. (b) The projectile returns to the ground after \(\boxed{5}\) seconds.