a projectile is fired from the surface of the earth with a velocity of `5ms^(-1)` and angle `theta` with the horizontal. Another projectile fired from another planet with a velocity of `3ms^(-1)` at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth.The value of the acceleration due to gravity on the planet is in `ms^(-2)` is given `(g=9.8 ms^(-2))`

Options

A. `3.5`
B. `5.9`
C. `16.3`
D. `110.8`

Explanation

To solve the problem, we need to compare the projectile motion of two projectiles fired from different planets but following the same trajectory.

### Step-by-Step Solution:

1. **Understand the Projectile Motion Equation**:
The trajectory of a projectile can be described by the equation:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
where:
- \(y\) is the height,
- \(x\) is the horizontal distance,
- \(g\) is the acceleration due to gravity,
- \(u\) is the initial velocity,
- \(\theta\) is the angle of projection.

2. **Set Up the Equations for Both Projectiles**:
For the projectile fired from Earth:
- Initial velocity, \(u_1 = 5 \, \text{m/s}\)
- Acceleration due to gravity, \(g_1 = 9.8 \, \text{m/s}^2\)

For the projectile fired from the other planet:
- Initial velocity, \(u_2 = 3 \, \text{m/s}\)
- Acceleration due to gravity, \(g_2\) (unknown)

3. **Equate the Trajectories**:
Since both projectiles follow the same trajectory, we can equate the terms involving gravity and initial velocity:
\[ \frac{g_1}{u_1^2} = \frac{g_2}{u_2^2} \]

4. **Substitute Known Values**:
Substitute the known values into the equation:
\[ \frac{9.8}{5^2} = \frac{g_2}{3^2} \]

5. **Calculate \(g_2\)**:
Simplifying the left side:
\[ \frac{9.8}{25} = \frac{g_2}{9} \]
Cross-multiplying gives:
\[ g_2 = 9.8 \times \frac{9}{25} \]
Now calculate:
\[ g_2 = \frac{88.2}{25} = 3.528 \, \text{m/s}^2 \]

6. **Final Answer**:
The acceleration due to gravity on the other planet is approximately:
\[ g_2 \approx 3.53 \, \text{m/s}^2 \]