A projectile is launched from ground level with an initial velocity of $v_{0}$ feet per second. Neglecting air resistance, its height in feet $t$ seconds after launch is given by $s=-16 t^{2}+v_{0} t$. Find the time(s) that the projectile will (a) reach a height of $64 \mathrm{ft}$ and $(b)$ return to the ground when $v_{0}=80$ feet per second.
(a) Find the time(s) that the projectile will reach a height of $64 \mathrm{ft}$ when $v_{0}=80$ feet per second. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. 1.0,4.0 seconds (Use a comma to separate answers as needed.)
B. The projectile does not reach 64 feet.
(b) The projectile returns to the ground after second(s).

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.