A ball is projected from ground with a speed `70m//s` at an angle `45^(@)` with the vertical so that it strikes a vertical wall at horizontal distance `(490)/(3)m` from the point of projection. If the ball returns back to the poin tof projection without any collision with ground then find the coefficient of restitution between the ball and wall.

Explanation

To solve the problem, we need to find the coefficient of restitution (e) between the ball and the wall after the ball is projected and strikes the wall. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the initial velocity components
The ball is projected with a speed \( u = 70 \, \text{m/s} \) at an angle \( \theta = 45^\circ \) with the vertical. We can find the horizontal and vertical components of the initial velocity:

- \( u_x = u \sin(45^\circ) = 70 \cdot \frac{1}{\sqrt{2}} = \frac{70}{\sqrt{2}} \, \text{m/s} \)
- \( u_y = u \cos(45^\circ) = 70 \cdot \frac{1}{\sqrt{2}} = \frac{70}{\sqrt{2}} \, \text{m/s} \)

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