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 will use the concept of projectile motion and the relationship between the maximum height of the projectiles fired from two different planets.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two projectiles:
- Projectile 1 is fired from Earth with an initial velocity \( u_e = 5 \, \text{m/s} \) at an angle \( \theta \).
- Projectile 2 is fired from another planet with an initial velocity \( u_p = 3 \, \text{m/s} \) at the same angle \( \theta \).
Both projectiles follow identical trajectories.

2. **Height of Projectile Formula**:
The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

3. **Height from Earth**:
For the projectile fired from Earth:
\[ H_e = \frac{u_e^2 \sin^2 \theta}{2g_e} \]
where \( g_e = 9.8 \, \text{m/s}^2 \).

4. **Height from the Planet**:
For the projectile fired from the other planet:
\[ H_p = \frac{u_p^2 \sin^2 \theta}{2g_p} \]
where \( g_p \) is the acceleration due to gravity on the planet.

5. **Setting the Heights Equal**:
Since the trajectories are identical, we can set the heights equal:
\[ H_e = H_p \]
This gives us:
\[ \frac{u_e^2 \sin^2 \theta}{2g_e} = \frac{u_p^2 \sin^2 \theta}{2g_p} \]

6. **Canceling Common Terms**:
We can cancel \( \sin^2 \theta \) and \( 2 \) from both sides:
\[ \frac{u_e^2}{g_e} = \frac{u_p^2}{g_p} \]

7. **Rearranging for \( g_p \)**:
Rearranging the equation gives us:
\[ g_p = \frac{u_p^2}{u_e^2} \cdot g_e \]

8. **Substituting Values**:
Now, substituting the values:
- \( u_p = 3 \, \text{m/s} \)
- \( u_e = 5 \, \text{m/s} \)
- \( g_e = 9.8 \, \text{m/s}^2 \)
\[ g_p = \frac{3^2}{5^2} \cdot 9.8 \]
\[ g_p = \frac{9}{25} \cdot 9.8 \]
\[ g_p = 0.36 \cdot 9.8 \]
\[ g_p = 3.528 \, \text{m/s}^2 \]

9. **Final Result**:
Rounding off, we find:
\[ g_p \approx 3.5 \, \text{m/s}^2 \]

### Final Answer:
The value of the acceleration due to gravity on the planet is approximately \( 3.5 \, \text{m/s}^2 \).

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