Question

Launched from the ground, a rocket accelerates vertically upward at 4.6 m/s2. It passes through a band of clouds 5.3 km thick, extending upward from an altitude of 1.9 km. How long is it in the clouds?

Answer 1

Answer on Question 55001, Physics, Mechanics | Kinematics | Dynamics

Question:

Launched from the ground, a rocket accelerates vertically upward at $4.6 \, \text{m} / \text{s}^2$. It passes through a band of clouds $5.3 \, \text{km}$ thick, extending upward from an altitude of $1.9 \, \text{km}$. How long is it in the clouds?

Solution:

Let's first find the time when the rocket enters the clouds. Because the initial velocity of the rocket $v_0 = 0$ (the rocket launched from rest) we can write:

$$y_1 = \frac{1}{2} a t_1^2,$$

where, $y_1$ is the height from which the rocket enters the clouds, $a$ is the acceleration of the rocket and $t_1$ is the time when the rocket enters the clouds.

From this formula we can calculate $t_1$:

$$t_1 = \sqrt{\frac{2 y_1}{a}} = \sqrt{\frac{2 \cdot 1.9 \cdot 10^3 \, \text{m}}{4.6 \, \frac{\text{m}}{\text{s}^2}}} = 28.7 \, \text{s}.$$

Then, we can find the time when the rocket leaves the clouds from the same formula:

$$y_2 = \frac{1}{2} a t_2^2,$$

where, $y_2$ is the height from which the rocket leaves the clouds, $a$ is the acceleration of the rocket and $t_2$ is the time when the rocket leaves the clouds.

So, we can calculate $t_2$:

$$t_2 = \sqrt{\frac{2 y_2}{a}} = \sqrt{\frac{2 \cdot (1.9 \cdot 10^3 \, \text{m} + 5.3 \cdot 10^3 \, \text{m})}{4.6 \, \frac{\text{m}}{\text{s}^2}}} = 56.0 \, \text{s}.$$

Finally, we can find the time spent in the clouds:

$$t = t_2 - t_1 = 56.0 \, \text{s} - 28.7 \, \text{s} = 27.3 \, \text{s}.$$

Answer: $t = 27.3 \, \text{s}$

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