If an object is projected upward from ground level with an initial velocity of 64 ft per sec, then its height in feet after I seconds is given by s(t)=-16t^2+64t Find the number of seconds it will take to reach its maximum height What is this maximum height? The object will take square second(s) to reach its maximum height. (Simplify your answer.) The maximum height reached by the object is square feet (Simplify your answer.)

To find the number of seconds it will take to reach the maximum height, we need to find the time at which the height is at its maximum. This occurs when the velocity of the object is zero.<br /><br />The velocity of the object is given by the derivative of the height function, $s(t)$. So, we need to find the derivative of $s(t)$ and set it equal to zero to find the time at which the velocity is zero.<br /><br />$s(t) = -16t^2 + 64t$<br /><br />Taking the derivative of $s(t)$ with respect to $t$, we get:<br /><br />$s'(t) = -32t + 64$<br /><br />Setting $s'(t)$ equal to zero and solving for $t$, we get:<br /><br />$-32t + 64 = 0$<br /><br />$-32t = -64$<br /><br />$t = 2$<br /><br />So, it will take 2 seconds for the object to reach its maximum height.<br /><br />To find the maximum height, we substitute $t = 2$ into the original height function, $s(t)$:<br /><br />$s(2) = -16(2)^2 + 64(2)$<br /><br />$s(2) = -64 + 128$<br /><br />$s(2) = 64$<br /><br />Therefore, the maximum height reached by the object is 64 feet.<br /><br />In summary:<br />The object will take $\boxed{2}$ second(s) to reach its maximum height.<br />The maximum height reached by the object is $\boxed{64}$ feet.