22. A rocket is shot straight into the air at an initial velocity of 30 feet/second. The height of the rocket after $t$ seconds is represented by the formula $h=30 t-5 t^{2}$. What is the maximum height the rocket reaches?
45 feet
65 feet
15 feet
55 feet

Step 1 ：The height of the rocket after $t$ seconds is represented by the formula \(h=30 t-5 t^{2}\).

Step 2 ：The maximum height of the rocket is achieved when the derivative of the height function is zero. This is because the derivative of a function at a point gives the slope of the tangent line at that point. When the derivative is zero, the tangent line is horizontal, which means the function has reached a maximum or minimum.

Step 3 ：In this case, we know it's a maximum because the coefficient of the \(t^2\) term is negative, which means the parabola opens downwards.

Step 4 ：Let's find the derivative of the height function: \(dh = 30 - 10*t\).

Step 5 ：Setting the derivative equal to zero, we find that \(t_{max} = 3\).

Step 6 ：Substituting \(t_{max}\) back into the height function, we find that the maximum height is \(h_{max} = 45\).

Step 7 ：Final Answer: The maximum height the rocket reaches is \(\boxed{45}\) feet.