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 $h=30t-5t^2$. 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 $h=30t-5t^2$.

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: $dh=30-10\ast t$.

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 $\boxed{{\displaystyle 45}}$ feet.