The acceleration due to gravity on a planet is `1.96 ms^(-1)`. If it is safe to jump from a height of 2 m on the earth, what will be the corresponding safe height on the planet?

Options

A. 5m
B. 2m
C. 10m
D. 20m

Answer

To solve the problem step by step, we will use the equations of motion to find the safe height from which a person can jump on a planet with a different acceleration due to gravity compared to Earth.

### Step 1: Understand the given data
- Acceleration due to gravity on Earth (gₑ) = 9.8 m/s²
- Safe jump height on Earth (hₑ) = 2 m
- Acceleration due to gravity on the planet (gₚ) = 1.96 m/s²

### Step 2: Find the final velocity (v) when jumping from 2 m on Earth
We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( u \) = initial velocity = 0 (since the person starts from rest)
- \( a \) = acceleration due to gravity on Earth = 9.8 m/s²
- \( s \) = height = 2 m

Substituting the values:
\[ v^2 = 0 + 2 \cdot 9.8 \cdot 2 \]
\[ v^2 = 39.2 \]
\[ v = \sqrt{39.2} \]
\[ v \approx 6.26 \, \text{m/s} \]

### Step 3: Use the final velocity to find the safe height on the planet
Now, we will use the same final velocity (v) to find the corresponding safe height (hₚ) on the planet with a different acceleration due to gravity:
Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( u \) = initial velocity = 0
- \( a \) = acceleration due to gravity on the planet = 1.96 m/s²
- \( s \) = height on the planet (hₚ) which we need to find

Substituting the values:
\[ v^2 = 0 + 2 \cdot 1.96 \cdot hₚ \]
Since we already found \( v^2 \approx 39.2 \):
\[ 39.2 = 2 \cdot 1.96 \cdot hₚ \]

### Step 4: Solve for hₚ
Rearranging the equation to find \( hₚ \):
\[ hₚ = \frac{39.2}{2 \cdot 1.96} \]
\[ hₚ = \frac{39.2}{3.92} \]
\[ hₚ = 10 \, \text{m} \]

### Final Answer
The corresponding safe height on the planet is **10 meters**.

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