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

Options

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

Answer

To solve the problem step by step, we will use the principles of kinematics and the relationship between height, gravity, and velocity.

### Step 1: Identify the known values
- Acceleration due to gravity on Earth, \( g_e = 9.81 \, \text{m/s}^2 \)
- Acceleration due to gravity on the planet, \( g_p = 1.96 \, \text{m/s}^2 \)
- Safe height on Earth, \( H_e = 3 \, \text{m} \)

### Step 2: Calculate the velocity just before hitting the ground on Earth
Using the kinematic equation:
\[ V^2 = U^2 + 2a s \]
where:
- \( V \) = final velocity
- \( U \) = initial velocity (0, since the person jumps)
- \( a \) = acceleration (gravity)
- \( s \) = height

For Earth:
\[ V_e^2 = 0 + 2 \cdot g_e \cdot H_e \]
Substituting the values:
\[ V_e^2 = 2 \cdot 9.81 \cdot 3 \]
Calculating:
\[ V_e^2 = 58.86 \quad \Rightarrow \quad V_e = \sqrt{58.86} \approx 7.67 \, \text{m/s} \]

### Step 3: Set up the equation for the planet
Using the same kinematic equation for the planet:
\[ V_p^2 = 2 \cdot g_p \cdot H_p \]
We want to find \( H_p \) when \( V_p = V_e \):
\[ V_e^2 = 2 \cdot g_p \cdot H_p \]
Substituting \( V_e^2 \) from the previous calculation:
\[ 58.86 = 2 \cdot 1.96 \cdot H_p \]

### Step 4: Solve for \( H_p \)
Rearranging the equation:
\[ H_p = \frac{58.86}{2 \cdot 1.96} \]
Calculating:
\[ H_p = \frac{58.86}{3.92} \approx 15.01 \, \text{m} \]

### Step 5: Conclusion
The corresponding height on the planet where it is safe to jump is approximately \( H_p \approx 15 \, \text{m} \).

### Final Answer
The corresponding height on the planet is **15 meters** (Option 4).

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