Solution 9: Rocket and Bolt
Given Data:
- Rocket acceleration: $a$ (upward, constant)
- Bolt dropped at: $t_1 = 1.0$ s
- Fuel finished at: $t_2 = 1.0 + 4.0 = 5.0$ s
- Air-time of bolt: $t_{air} = 2.0$ s
- $g = 10$ m/s²
Step 1: Analyze Bolt Motion to find Acceleration ($a$)
At $t=1$, the rocket has:
$$ v_{rocket} = a \times 1 = a $$
$$ h_{rocket} = \frac{1}{2} a (1)^2 = 0.5a $$
The bolt starts with initial upward velocity $u = a$ from height $h = 0.5a$.
Using displacement equation $S = ut – \frac{1}{2}gt^2$ for the bolt (where $S = -h$):
$$ -0.5a = a(2) – \frac{1}{2}(10)(2)^2 $$
$$ -0.5a = 2a – 20 $$
$$ 2.5a = 20 \implies a = 8.0 \text{ m/s}^2 $$
Statement (a) is correct.
Step 2: Rocket Height when Fuel Finishes
Fuel finishes at $t = 1 + 4 = 5$ s.
$$ h_{fuel} = \frac{1}{2} a t^2 = \frac{1}{2} (8) (5)^2 $$
$$ h_{fuel} = 4 \times 25 = 100 \text{ m} $$
Statement (b) is correct.
Step 3: Maximum Speed of Rocket
The rocket accelerates until fuel runs out ($t=5$). Afterward, it moves under gravity, so speed decreases.
$$ v_{max} = a \times t = 8 \times 5 = 40 \text{ m/s} $$
Statement (c) is correct.
Step 4: Total Air-time of Rocket
After fuel finishes ($t=5$), the rocket is at $h=100$ m with $u=40$ m/s (upward). It is now in free fall.
Let $t’$ be the time to hit ground from this point ($S = -100$).
$$ -100 = 40t’ – \frac{1}{2}(10)(t’)^2 $$
$$ -100 = 40t’ – 5(t’)^2 $$
$$ 5(t’)^2 – 40t’ – 100 = 0 $$
$$ (t’)^2 – 8t’ – 20 = 0 $$
$$ (t’ – 10)(t’ + 2) = 0 $$
Time must be positive, so $t’ = 10$ s.
Total time = Time with fuel + Free fall time = $5 + 10 = 15$ s.
Statement (d) is correct.
Answer: (a), (b), (c), and (d) are all correct.