A ball is thrown from the ground so that it just crosses a wall 5m high at a distance of 10m and falls at a distance of 10m ahead from the wall. Find the speed and the direction of projection of ball. Assume, `g = 10" ms"^(–2)`

Explanation

To solve the problem of a ball thrown from the ground that just crosses a wall 5 meters high at a distance of 10 meters, and then falls 10 meters ahead of the wall, we can break down the solution into clear steps.

### Step 1: Understand the Problem
We need to find the initial speed (U) and the angle of projection (θ) of the ball. The wall height (H) is 5 m, the distance from the point of projection to the wall (d1) is 10 m, and the total horizontal distance from the point of projection to where the ball lands (d_total) is 20 m.

### Step 2: Use the Maximum Height Formula
The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{U^2 \sin^2 \theta}{2g} \]
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