(1 point)
A model rocket is launched with an initial velocity of $168 \mathrm{ft} / \mathrm{s}$ from a height of $30 \mathrm{ft}$. The height of the rocket, in feet, $t$ seconds after it has been launched is given by the function $s(t)=-16 t^{2}+168 t+30$. Determine the time which the rocket reaches its maximum height and find the maximum height.
Time to Reach Max Height:
s
Maximum Height:
$\mathrm{ft}$
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Step 1 ：The maximum height of a projectile launched upward like a rocket is achieved when its velocity is 0. In this case, the velocity of the rocket is given by the derivative of the height function, s(t).

Step 2 ：So, to find the time at which the rocket reaches its maximum height, we need to set the derivative of s(t) to 0 and solve for t.

Step 3 ：The maximum height is then found by substituting this time into the height function, s(t).

Step 4 ：Let's denote the height function as \(s(t) = -16t^2 + 168t + 30\).

Step 5 ：The derivative of the height function, \(s'(t)\), is \(168 - 32t\).

Step 6 ：Setting \(s'(t)\) to 0, we get \(t = \frac{21}{4}\).

Step 7 ：Substituting \(t = \frac{21}{4}\) into the height function, we get the maximum height, \(s_{max} = 471\).

Step 8 ：Final Answer: The time at which the rocket reaches its maximum height is \(\boxed{\frac{21}{4}}\) seconds and the maximum height is \(\boxed{471}\) feet.