A toy gun a spring of force constant `k`. When changed before being triggered in the upward direction, the spring is compressed by a distance `x`. If the mass of the shot is `m`, on the being triggered it will go up to a height of

Options

A. `(kx^(2))/(mg)`
B. `(x^(2))/(kmg)`
C. `(kx^(2))/(2mg)`
D. `((kx)^(2))/(mg)`

Explanation

To solve the problem, we will use the principle of conservation of energy. The elastic potential energy stored in the spring when it is compressed will be converted into gravitational potential energy when the mass is shot upwards.

### Step-by-Step Solution:

1. **Identify the Elastic Potential Energy (EPE) of the Spring:**
The elastic potential energy stored in a compressed spring is given by the formula:
\[ EPE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the compression distance.

2. **Identify the Gravitational Potential Energy (GPE):**
When the mass \( m \) is shot upwards to a height \( h \), the gravitational potential energy gained by the mass is given by:
\[ GPE = mgh \]
where \( g \) is the acceleration due to gravity.

3. **Apply the Conservation of Energy Principle:**
According to the conservation of energy, the elastic potential energy stored in the spring will be equal to the gravitational potential energy gained by the mass when it is shot upwards:
\[ \frac{1}{2} k x^2 = mgh \]

4. **Solve for Height \( h \):**
Rearranging the equation to solve for \( h \):
\[ h = \frac{\frac{1}{2} k x^2}{mg} \]
Simplifying this gives:
\[ h = \frac{k x^2}{2mg} \]

5. **Final Answer:**
Therefore, the height \( h \) to which the mass will rise is:
\[ h = \frac{k x^2}{2mg} \]