The acceleration due to gravity on a planet is `1.96 m//s^(2)`. If a boy can safely jump from a height of 2 m on the earth, then the coorresponding safe height on the planet will be

Options

A. 2.5 m
B. 5 m
C. 7.5 m
D. 10 m

Explanation

To solve the problem of determining the safe jumping height on a planet where the acceleration due to gravity is \(1.96 \, \text{m/s}^2\), given that a boy can safely jump from a height of \(2 \, \text{m}\) on Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the relationship between height and gravity
The potential energy (PE) at a height \(h\) is given by the formula:
\[ PE = mgh \]
where \(m\) is the mass of the boy, \(g\) is the acceleration due to gravity, and \(h\) is the height.

### Step 2: Calculate the velocity just before landing on Earth
When the boy jumps from a height \(h\) on Earth, the potential energy at that height is converted to kinetic energy (KE) just before he lands. The kinetic energy is given by:
\[ KE = \frac{1}{2} mv^2 \]
At the height of \(2 \, \text{m}\) on Earth, we have:
\[ mgh = \frac{1}{2} mv^2 \]
Cancelling \(m\) from both sides (assuming \(m \neq 0\)):
\[ gh = \frac{1}{2} v^2 \]
Rearranging gives:
\[ v^2 = 2gh \]
Substituting \(g = 10 \, \text{m/s}^2\) (approximate value for Earth) and \(h = 2 \, \text{m}\):
\[ v^2 = 2 \times 10 \times 2 = 40 \]
Thus, the velocity just before landing on Earth is:
\[ v = \sqrt{40} = 6.32 \, \text{m/s} \]

### Step 3: Set up the equation for the planet
On the planet, we want to find the height \(h'\) from which the boy can jump safely. The same principle applies:
\[ mgh' = \frac{1}{2} mv'^2 \]
where \(g' = 1.96 \, \text{m/s}^2\) is the acceleration due to gravity on the planet. The velocity just before landing on the planet will also be:
\[ v' = \sqrt{2g'h'} \]

### Step 4: Equate the velocities
Since the boy can safely jump from the same height on both Earth and the planet, we can equate the velocities:
\[ \sqrt{2gh} = \sqrt{2g'h'} \]
Squaring both sides gives:
\[ 2gh = 2g'h' \]
Cancelling \(2\) from both sides:
\[ gh = g'h' \]

### Step 5: Solve for \(h'\)
Rearranging the equation gives:
\[ h' = \frac{gh}{g'} \]
Substituting \(g = 10 \, \text{m/s}^2\), \(h = 2 \, \text{m}\), and \(g' = 1.96 \, \text{m/s}^2\):
\[ h' = \frac{10 \times 2}{1.96} \approx \frac{20}{1.96} \approx 10.2 \, \text{m} \]

### Conclusion
Thus, the corresponding safe height on the planet is approximately \(10.2 \, \text{m}\).