2. If projectile is thrown at an angle (X) with horizontal with ground towards a wall 4m away from point of projection and height is 4m this projectile just touches ,thrown on other side at 14m? ?

To solve this problem, we can break it down into two separate projectile motion problems - one for the initial throw towards the wall and one for the throw on the other side.

1. Initial throw towards the wall:
Let's assume the initial velocity of the projectile is V m/s and the angle of projection is X degrees. The horizontal distance to the wall is 4m and the height is 4m.

Using the equations of motion for projectile motion, we can calculate the time taken for the projectile to reach the wall:
Horizontal distance = V*cos(X)*t
4 = V*cos(X)*t

Vertical distance = V*sin(X)*t - (1/2)*g*t^2
4 = V*sin(X)*t - (1/2)*9.8*t^2

Solving these two equations simultaneously, we can find the values of V and X.

2. Throw on the other side:
Now, the projectile is thrown on the other side at a distance of 14m. We can use the same equations of motion to calculate the angle of projection and initial velocity required for the projectile to just touch the wall on the other side.

Horizontal distance = V*cos(X)*t
14 = V*cos(X)*t

Vertical distance = V*sin(X)*t - (1/2)*g*t^2
4 = V*sin(X)*t - (1/2)*9.8*t^2

Solving these two equations simultaneously, we can find the angle of projection and initial velocity required for the projectile to just touch the wall on the other side.

I hope this helps! Let me know if you need further clarification.