A small ball is thrown from the edge of one bank of a river of width `100m` to just reach the bank. The ball was thrown in the vertical plane (which is also perpendicular to the banks) at an angle `37^(@)` to the horizonatal. Taking the starting the point as the origin `O`, verection as positeve `y`-axis and horizonrtal line passing through the point `O` and perpenducular to the bank as `x`-axis find:
(a) Equation of trahectory of the image formed by the water surface
(water surface is at the level `y=0`)
(b) instantaneous velocity of the image formed due to refraction. [Use `g=10 m//s^(2),R.I.` of water `=4//3`]

Explanation

To solve the problem step by step, we will break it down into two parts as requested.

### Part (a): Equation of the trajectory of the image formed by the water surface

1. **Understanding the trajectory of the ball**:
The ball is thrown at an angle of \(37^\circ\) to the horizontal. The trajectory of the ball in the absence of refraction can be described by the equation:
\[ y = x \tan(\theta) - \frac{g}{2u^2 \cos^2(\theta)} x^2 \]
where \(g\) is the acceleration due to gravity, \(u\) is the initial velocity, and \(\theta\) is the angle of projection.

2. **Using the given values**:
We know the width of the river \(R = 100 \, m\) and the angle \(\theta = 37^\circ\). The tangent of \(37^\circ\) is \(\tan(37^\circ) = \frac{3}{4}\).

3. **Finding the equation of the actual trajectory**:
The trajectory of the ball can be simplified to:
\[ y = x \cdot \frac{3}{4} - \frac{10}{2u^2 \cdot \left(\frac{4}{5}\right)^2} x^2 \]
where \(g = 10 \, m/s^2\) and \(\cos(37^\circ) = \frac{4}{5}\).

4. **Finding the image trajectory due to refraction**:
The trajectory of the image formed by the water surface can be obtained by multiplying the \(y\)-coordinate of the actual trajectory by the refractive index of water, which is \(\frac{4}{3}\):
\[ y' = \frac{4}{3}y \]

5. **Substituting the trajectory equation**:
Substituting the trajectory equation into the image trajectory equation, we have:
\[ y' = \frac{4}{3} \left( x \cdot \frac{3}{4} - \frac{10}{2u^2 \cdot \left(\frac{4}{5}\right)^2} x^2 \right) \]
Simplifying this gives:
\[ y' = x - \frac{4}{3} \cdot \frac{10}{2u^2 \cdot \left(\frac{16}{25}\right)} x^2 \]
Thus, the equation of the trajectory of the image is:
\[ y' = x - \frac{5}{3} \cdot \frac{10}{u^2} x^2 \]

### Part (b): Instantaneous velocity of the image formed due to refraction

1. **Finding the initial velocity \(u\)**:
The range \(R\) of the projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Rearranging for \(u^2\):
\[ u^2 = \frac{R \cdot g}{\sin(2\theta)} \]
With \(R = 100 \, m\), \(g = 10 \, m/s^2\), and \(\sin(2 \cdot 37^\circ) = 2 \cdot \sin(37^\circ) \cdot \cos(37^\circ) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}\):
\[ u^2 = \frac{100 \cdot 10}{\frac{24}{25}} = \frac{2500}{24} \Rightarrow u \approx 20.41 \, m/s \]

2. **Finding the instantaneous velocity**:
The instantaneous velocity \(v\) of the ball at any point can be expressed in vector form as:
\[ v = v_x \hat{i} + v_y \hat{j} \]
where \(v_x = u \cos(\theta)\) and \(v_y = u \sin(\theta) - gt\).

3. **Calculating components**:
Using \(u \approx 20.41 \, m/s\), \(\cos(37^\circ) = \frac{4}{5}\), and \(\sin(37^\circ) = \frac{3}{5}\):
\[ v_x = 20.41 \cdot \frac{4}{5} \approx 16.33 \, m/s \]
\[ v_y = 20.41 \cdot \frac{3}{5} - 10t \]

4. **Considering refraction**:
The \(y\) component of the velocity of the image will be multiplied by the refractive index:
\[ v_y' = \frac{4}{3} v_y \]

5. **Final expression for instantaneous velocity**:
The instantaneous velocity of the image is:
\[ v' = v_x \hat{i} + \frac{4}{3}(20.41 \cdot \frac{3}{5} - 10t) \hat{j} \]