A rocket is fired upward from the earth's surface such that it creates an acceleration of 19.6 m/sec . If after 5 sec its engine is switched off, the maximum height of the rocket from earth's surface would be

Options

A. 245 m
B. 490 m
C. 980 m
D. 735 m

Explanation

To solve the problem of finding the maximum height of a rocket fired upward from the Earth's surface, we can break it down into several steps.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Acceleration of the rocket, \( a = 19.6 \, \text{m/s}^2 \)
- Time of engine firing, \( t = 5 \, \text{s} \)
- Initial velocity, \( u = 0 \, \text{m/s} \) (since the rocket starts from rest)

2. **Calculate Final Velocity After 5 Seconds:**
We can use the equation of motion:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + (19.6 \, \text{m/s}^2)(5 \, \text{s}) = 98 \, \text{m/s} \]

3. **Calculate the Distance Covered During the First 5 Seconds:**
We can use the equation:
\[ x = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ x = 0 \cdot 5 + \frac{1}{2} (19.6 \, \text{m/s}^2)(5 \, \text{s})^2 \]
\[ x = \frac{1}{2} (19.6)(25) = 245 \, \text{m} \]

4. **Calculate the Maximum Height After the Engine is Turned Off:**
After the engine is turned off, the rocket will continue to rise until its velocity becomes zero. We can use the following equation:
\[ v^2 = u^2 - 2gh \]
Here, \( v = 0 \) (at maximum height), \( u = 98 \, \text{m/s} \), and \( g = 9.8 \, \text{m/s}^2 \). Rearranging for \( h \):
\[ 0 = (98)^2 - 2(9.8)h \]
\[ 2(9.8)h = (98)^2 \]
\[ h = \frac{(98)^2}{2 \cdot 9.8} = \frac{9604}{19.6} = 490 \, \text{m} \]

5. **Calculate the Total Maximum Height:**
The total height \( H \) is the sum of the distance covered during the first 5 seconds and the additional height gained after the engine is turned off:
\[ H = x + h = 245 \, \text{m} + 490 \, \text{m} = 735 \, \text{m} \]

### Final Answer:
The maximum height of the rocket from the Earth's surface is **735 meters**.

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