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=-16t^2+v_(0)t Find the time(s) that the projectile will (a)reach a heig of 288 ft and (b) return to the ground when v_(0)=144 feet per second. (a) Find the time(s)that the projectile will reach a height of 288 ft when v_(0)=144 feet per second. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. square seconds (Use a comma to separate answers as needed.) B. The projectile does not reach 288 feet.

To find the time(s) that the projectile will reach a height of 288 feet, we need to solve the equation $s = -16t^2 + v_0t$ for $t$ when $s = 288$ and $v_0 = 144$.<br /><br />Substituting the given values into the equation, we have:<br /><br />$288 = -16t^2 + 144t$<br /><br />Rearranging the equation, we get:<br /><br />$-16t^2 + 144t - 288 = 0$<br /><br />Dividing both sides of the equation by -16, we have:<br /><br />$t^2 - 9t + 18 = 0$<br /><br />Now, we can solve this quadratic equation using factoring or the quadratic formula. Let's try factoring first:<br /><br />$t^2 - 9t + 18 = (t - 3)(t - 6) = 0$<br /><br />Setting each factor equal to zero, we have:<br /><br />$t - 3 = 0$ or $t - 6 = 0$<br /><br />Solving for $t$, we get:<br /><br />$t = 3$ or $t = 6$<br /><br />Therefore, the correct answer is:<br /><br />A. 3 seconds, 6 seconds