A projectile is launched from ground level with an initial velocity of y_(0) feel pen second. Neglecting sir resistance, its height in feel I seconds short tourch is given by s=-16t^2+v_(0)t Find the time(s) that the projectle will (8) reach a height of 80 it and (b) return to the ground when vo feel per swared. (a) Find the time(s)that the projectle will reach a height of 80 it; when v_(0)=96 leaf per second. Select the connect choice below and,if necessary, I'd in the answer box to complete your choice. (A) A. square seconds (Use a comma to separate answers at neededs OB. The projectile does not reach 50 feet.

To find the time(s) that the projectile will reach a height of 80 feet, we need to solve the equation $s = -16t^2 + v_0t$ for $t$ when $s = 80$.<br /><br />Given that $v_0 = 96$ feet per second, we can substitute these values into the equation:<br /><br />$80 = -16t^2 + 96t$<br /><br />Rearranging the equation, we have:<br /><br />$-16t^2 + 96t - 80 = 0$<br /><br />This is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = -16$, $b = 96$, and $c = -80$. We can solve this equation using the quadratic formula:<br /><br />$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />Plugging in the values, we get:<br /><br />$t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(-80)}}{2(-16)}$<br /><br />Simplifying further, we have:<br /><br />$t = \frac{-96 \pm \sqrt{9216 - 5120}}{-32}$<br /><br />$t = \frac{-96 \pm \sqrt{4096}}{-32}$<br /><br />$t = \frac{-96 \pm 64}{-32}$<br /><br />This gives us two possible solutions for $t$:<br /><br />$t_1 = \frac{-96 + 64}{-32} = 1$<br /><br />$t_2 = \frac{-96 - 64}{-32} = 5$<br /><br />Therefore, the projectile will reach a height of 80 feet at $t = 1$ second and $t = 5$ seconds.<br /><br />So, the correct answer is (A) 1, 5 seconds.