Give step-by-step solution with explanation and final answer:
A small rocket is launched vertically upward. It maintains a constant acceleration of 6 m/s2 until its fuel is fully consumed 20 seconds a

Question

Give step-by-step solution with explanation and final answer:
A small rocket is launched vertically upward. It maintains a constant acceleration of 6 m/s2 until its fuel is fully consumed 20 seconds after launch. Find: The speed reached by the rocket at the instant the fuel is fully consumed, The maximum height reached by the rocket, The total time the rocket is in the air, and The speed of the rocket when it hits the ground at the end of its flight

VIPSolutions · Accepted Answer

# Rocket Motion Problem Solution

## Given Data

- **Upward acceleration** (while fuel burns): $ a = 6\,\mathrm{m/s^2} $
- **Time of powered flight:** $ t_1 = 20\,\mathrm{s} $
- **Initial velocity:** $ u =  $
- **Acceleration due to gravity:** $ g = 9.8\,\mathrm{m/s^2} $ (downward)

---

## 1. Speed When Fuel is Consumed

The rocket accelerates upward for $ t_1 = 20\,\mathrm{s} $:

$$
v_1 = u + a t_1
$$
$$
v_1 =  + (6)(20) = 120\,\mathrm{m/s}
$$

**Final speed after fuel is burned:**  
$$
\boxed{120\,\mathrm{m/s}}
$$

---

## 2. Maximum Height Reached

There are two phases:
1. **Powered flight** ($ t_1 = 20\,\mathrm{s} $) with $ a = 6\,\mathrm{m/s^2} $
2. **Free flight** (after fuel runs out, $ a = -g $), starting upward with $ v_1 = 120\,\mathrm{m/s} $

### a) Height after Powered Flight

$$
h_1 = ut_1 + \frac{1}{2} a t_1^2
$$
$$
h_1 =  + \frac{1}{2} (6)(20^2) = 3 \times 400 = 120\,\mathrm{m}
$$

### b) Additional Height During Free Flight

After fuel is used, the rocket rises until velocity is zero:

$$
v = v_1 - g t_2 = 
$$
$$
t_2 = \frac{v_1}{g} = \frac{120}{9.8} \approx 12.24\,\mathrm{s}
$$

Height gained during this time:

$$
h_2 = v_1 t_2 - \frac{1}{2} g t_2^2
$$
$$
h_2 = (120)(12.24) - \frac{1}{2}(9.8)(12.24^2)
$$
$$
h_2 = 1468.8 - .5 \times 9.8 \times 149.78
$$
$$
= 1468.8 - 4.9 \times 149.78 \approx 1468.8 - 734
$$
$$
\approx 734.8\,\mathrm{m}
$$

### c) Maximum Height

$$
H_{max} = h_1 + h_2 = 120 + 734.8 = 1934.8\,\mathrm{m}
$$

**Maximum height reached:**  
$$
\boxed{1935\,\mathrm{m}}
$$

---

## 3. Total Time in Air

### a) Time to Rise to Maximum Height

$$
t_{\text{up}} = t_1 + t_2 = 20 + 12.24 = 32.24\,\mathrm{s}
$$

### b) Time to Fall from Maximum Height

Fall from rest (initial velocity ) from $ H_{max} $:

$$
H_{max} = \frac{1}{2} g t_3^2
$$
$$
t_3 = \sqrt{\frac{2 H_{max}}{g}} = \sqrt{\frac{2 \times 1934.8}{9.8}}
$$
$$
= \sqrt{394.96} \approx 19.88\,\mathrm{s}
$$

### c) Total Time

$$
t_{\text{total}} = t_1 + t_2 + t_3 = 20 + 12.24 + 19.88 \approx 52.1\,\mathrm{s}
$$

**Total time in the air:**  
$$
\boxed{52.1\,\mathrm{s}}
$$

---

## 4. Speed When Rocket Hits the Ground

Falling from rest from $ H_{max} $:

$$
v = \sqrt{2 g H_{max}}
$$
$$
v = \sqrt{2 \times 9.8 \times 1934.8}
$$
$$
= \sqrt{37,926.08} \approx 194.8\,\mathrm{m/s}
$$

**Speed when hitting the ground:**  
$$
\boxed{194.8\,\mathrm{m/s}}
$$

---

## **Summary Table**

| Quantity                        | Value              |
|----------------------------------|--------------------|
| Speed after fuel burns           | $120\,\mathrm{m/s}$      |
| Maximum height                   | $1935\,\mathrm{m}$        |
| Total time in air                | $52.1\,\mathrm{s}$        |
| Speed on impact with ground      | $194.8\,\mathrm{m/s}$     |

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