In a different planet, the acceleration due to gravity is `2m//s^(2)`. A particle is thrown horizontally at a speed of `3 m//s` from a height of `10 m`. The equation of trajectory of the particle comes out to be `x^(2)=alphay`. Find the value of `alpha` [Take `+X-`axis alonf the direction of projection and `+Y` axis towards the surface of planet.]

Options

A. `7`
B. `8`
C. `9`
D. `10`

Explanation

To solve the problem, we need to find the value of α in the equation of trajectory \( x^2 = \alpha y \) for a particle thrown horizontally from a height of 10 m with an initial horizontal speed of 3 m/s and under the influence of gravity of 2 m/s².

### Step-by-Step Solution:

1. **Identify the parameters:**
- Initial height (h) = 10 m
- Initial horizontal speed (u_x) = 3 m/s
- Acceleration due to gravity (g) = 2 m/s²

2. **Determine the time of flight (t):**
The vertical motion can be described by the equation:
\[ y = h - \frac{1}{2} g t^2 \]
Here, since the particle is thrown from a height of 10 m, we can set \( y = 0 \) when it reaches the ground.
\[ 0 = 10 - \frac{1}{2} \cdot 2 \cdot t^2 \]
Simplifying this gives:
\[ 0 = 10 - t^2 \]
\[ t^2 = 10 \implies t = \sqrt{10} \]

3. **Calculate the horizontal distance (x):**
The horizontal distance traveled by the particle can be calculated using:
\[ x = u_x \cdot t \]
Substituting the values:
\[ x = 3 \cdot \sqrt{10} \]

4. **Relate x and y:**
We can express y in terms of x. From the vertical motion equation:
\[ y = h - \frac{1}{2} g t^2 \]
We already found that \( t^2 = 10 \), so:
\[ y = 10 - \frac{1}{2} \cdot 2 \cdot 10 = 0 \]
However, we need to express y in terms of x. We can use the time \( t \) in the horizontal motion:
\[ t = \frac{x}{u_x} = \frac{x}{3} \]
Substitute this into the vertical motion equation:
\[ y = 10 - \frac{1}{2} \cdot 2 \left(\frac{x}{3}\right)^2 \]
Simplifying gives:
\[ y = 10 - \frac{1}{2} \cdot 2 \cdot \frac{x^2}{9} = 10 - \frac{x^2}{9} \]

5. **Rearranging the equation:**
Rearranging the equation \( y = 10 - \frac{x^2}{9} \) gives:
\[ \frac{x^2}{9} = 10 - y \]
Multiplying through by -1:
\[ x^2 = 9(10 - y) \]
Rearranging gives:
\[ x^2 = -9y + 90 \]
This can be rewritten as:
\[ x^2 = 9(10 - y) \]
Thus, we can express it in the form \( x^2 = \alpha y \) where \( \alpha = 9 \).

6. **Final result:**
Therefore, the value of \( \alpha \) is:
\[ \alpha = 9 \]