A projectile is launched from ground level with an initial velocity of v_(0) feet per second. Neglecting air resistance, its height in feet I seconds after launch is given by s=-16t^2+v_(0)t Find the time(s) that the projectile will (a)reach a height of 80 ft and (b)return to the ground when v_(0)=96 feet per second. (a) Find the time(s)that the projectile will reach a height of 80 ft when v_(0)=96 feet per second. Select the correct choice below and,If necessary, fill in the answer box to complete your choice 1,5 seconds (Use a comma to separate answers as needed.) B. The projectile does not reach 80 feet. (b) The projectile retums to the ground after square second(s).

## Step 1<br />We are given the equation for the height of the projectile as \(s=-16t^{2}+v_{0}t\). We are asked to find the time(s) when the projectile reaches a height of 80ft and when it returns to the ground.<br /><br />## Step 2<br />For part (a), we need to find the time(s) when the projectile reaches a height of 80ft. We can do this by setting the equation equal to 80 and solving for \(t\).<br /><br />### \(-16t^{2}+v_{0}t = 80 \)<br /><br />## Step 3<br />For part (b), we need to find the time when the projectile returns to the ground. This is when the height \(s\) is 0. We can find this by setting the equation equal to 0 and solving for \(t\).<br /><br />### \( -16t^{2}+v_{0}t = 0 \)<br /><br />## Step 4<br />We are given that the initial velocity \(v_{0}\) is 96 feet per second. We can substitute this value into the equations from steps 2 and 3 to find the solutions.