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.