The accelearation due to gravity on a planet is 1.96 `ms^(-2)` if tit is safe to jump from a height of 3 m on the earth the corresponding height on the planet will be

Options

A. 3 m
B. 6m
C. 9m
D. 15m

Explanation

To solve the problem, we need to find the corresponding height on the planet where the acceleration due to gravity is 1.96 m/s², given that it is safe to jump from a height of 3 m on Earth where the acceleration due to gravity is approximately 9.8 m/s².

### Step-by-Step Solution:

1. **Understand the Problem**:
We know that a person can jump from a height of 3 m on Earth safely. We need to find out what height would be safe to jump from on another planet where the acceleration due to gravity is 1.96 m/s².

2. **Use the Kinematic Equation**:
We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2gh \]
where:
- \( v \) = final velocity (just before hitting the ground)
- \( u \) = initial velocity (0 m/s when jumping)
- \( g \) = acceleration due to gravity
- \( h \) = height jumped

3. **Calculate Final Velocity on Earth**:
For the jump on Earth:
- \( u = 0 \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 3 \, \text{m} \)

Plugging these values into the equation:
\[ v^2 = 0 + 2 \times 9.8 \times 3 \]
\[ v^2 = 58.8 \]
\[ v = \sqrt{58.8} \approx 7.67 \, \text{m/s} \]

4. **Use the Final Velocity to Find Height on the Planet**:
Now, we will use the same final velocity to find the corresponding height on the planet where \( g = 1.96 \, \text{m/s}^2 \):
\[ v^2 = u^2 + 2gh \]
Here, \( u = 0 \) and \( g = 1.96 \, \text{m/s}^2 \):
\[ (7.67)^2 = 0 + 2 \times 1.96 \times h \]
\[ 58.8 = 3.92h \]
\[ h = \frac{58.8}{3.92} \approx 15 \, \text{m} \]

5. **Conclusion**:
The corresponding height on the planet from which it is safe to jump is approximately **15 m**.