Question

A model rocket fired from the ground ascends with a constant upward acceleration. A small bolt is dropped from the rocket 1.0 s after the firing and fuel of the rocket is finished 4.0 s after the bolt is dropped. Air-time of the bolt is 2.0 s. Acceleration of free fall is 10 m/s2. Which of the following statements is/are correct?

• A. Acceleration of the rocket while ascending on its fuel is 8.0 m/s2
• B. Fuel of the rocket was finished at a height 100 m above the ground.
• C. Maximum speed of the rocket during its upward flight is 40 m/s.
• D. Total air-time of the rocket is 15 s.

Answer 1

Answer: (A), (B), (C), (D)

All the statements (a), (b), (c), and (d) are correct.

Answer 2

Solution Explanation:

1. Let the net upward acceleration be $a$. When the bolt is dropped at $t = 1$ s, its upward speed is $v_1 = a\cdot1 = a$ and height $h_1 = \frac{1}{2}a$.
2. The bolt then falls under gravity ($g = 10\,\text{m/s}^2$) and its displacement in 2 s is
     $\Delta h = a\cdot2 - \frac{1}{2}\times10\times2^2 = 2a - 20.$
   Since the bolt reaches the ground:
     $h_1 + (2a - 20) = 0 \quad\Longrightarrow\quad \frac{1}{2}a + 2a = 20,$   $2.5a = 20 \quad\Longrightarrow\quad a = \frac{20}{2.5} = 8\,\text{m/s}^2.$
3. Fuel runs out at $t = 1 + 4 = 5$ s. Height at burnout:
     $h = \frac{1}{2}\times8\times5^2 = 100\,\text{m}.$
4. Rocket’s speed at burnout:
     $v = 8\times5 = 40\,\text{m/s}.$
5. After burnout, the rocket continues upward as a projectile. Additional time to reach maximum height:
     $t = \frac{40}{10} = 4\,\text{s},$   and the rocket takes $t = \sqrt{\frac{2\times180}{10}} = 6\,\text{s}$ to fall from the maximum height (since maximum height = $100 + \frac{40^2}{2\times10} = 180\,\text{m}$).
6. Total air-time of the rocket is:
     $5\,\text{s} (\text{powered}) + 4\,\text{s}(\text{coasting up}) + 6\,\text{s}(\text{falling}) = 15\,\text{s}.$