5. A rocket is launched upwards from the top of a $30 \mathrm{~m}$ building, eventually falling to the ground below. The motion of the rocket is given by the equation, $\mathrm{h}=-5 \mathrm{t}^{2}+25 \mathrm{t}+30$ where h represents the height (in metres) of the rocket at a given time, $t$ (in seconds).
a) For how long is the rocket in the air?
Answer
Since the rocket cannot be in the air for a negative amount of time, we discard the negative solution. Therefore, the rocket is in the air for \(\boxed{6}\) seconds.
Step 1 ：The rocket is in the air from the time it is launched until it hits the ground. The time when it hits the ground is when the height, h, is zero. So, we need to solve the equation \(-5t^2 + 25t + 30 = 0\) for t.
Step 2 ：This is a quadratic equation, and we can solve it using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a, b, and c are the coefficients of the quadratic equation.
Step 3 ：In this case, a = -5, b = 25, and c = 30. We are interested in the larger of the two solutions, because the smaller solution represents the time before the rocket is launched.
Step 4 ：By substituting the values of a, b, and c into the quadratic formula, we get two solutions for t: t1 = -1.0 and t2 = 6.0.
Step 5 ：Since the rocket cannot be in the air for a negative amount of time, we discard the negative solution. Therefore, the rocket is in the air for \(\boxed{6}\) seconds.
