32. The initial velocity of a rocket shot straight up is 60 feet/second. The height of the rocket after $t$ seconds is represented by the equation $d=60 t-5 t^{2}$. Find the maximum height of the rocket and the amount of time the rocket is in the air. Maximum height = feet Total time in the air = seconds

Step 1 ：The height of the rocket is given by a quadratic equation. The maximum height of the rocket can be found by finding the vertex of the parabola represented by the equation. The vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is at the point \(-b/2a, f(-b/2a)\). In this case, \(a = -5\) and \(b = 60\), so the time at which the maximum height is reached is \(-b/2a = -60/-10 = 6\) seconds. We can substitute this value into the equation to find the maximum height.

Step 2 ：The rocket is in the air as long as its height is greater than 0. This is the case when \(60t - 5t^2 > 0\). We can solve this inequality to find the total time the rocket is in the air.

Step 3 ：Substitute \(t = 6.0\) into the equation \(height = -5*t**2 + 60*t\) to get the maximum height \(180.000000000000\) feet.

Step 4 ：Solve the equation \(60t - 5t^2 = 0\) to get the total time the rocket is in the air \([0, 12]\) seconds.

Step 5 ：Final Answer: The maximum height of the rocket is \(\boxed{180}\) feet and the total time the rocket is in the air is \(\boxed{12}\) seconds.