A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is :

A ball is projected from the ground at an angle of 45°. The point of projection is 4 m from the foot of the wall, and the ball lands 6 m beyond the wall. The total horizontal distance covered (range) is the sum of these distances: 4 m + 6 m = 10 m.

For a projectile launched at an angle θ with initial velocity u, the range R is given by:

$$R = \frac{u^2 \sin 2\theta}{g} $$

Here, θ = 45°, so 2θ = 90° and sin 90° = 1. Thus:

$$R = \frac{u^2}{g} $$

Given R = 10 m:

$$10 = \frac{u^2}{g} $$

Solving for u²:

$$u^2 = 10g \quad \text{(Equation 1)} $$

The trajectory equation for a projectile is:

$$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} $$

With θ = 45°, tan 45° = 1 and cos 45° = $$\frac{1}{\sqrt{2}}$$, so cos² 45° = $$\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$$. Substituting these values:

$$y = x \cdot 1 - \frac{g x^2}{2 u^2 \cdot \frac{1}{2}} $$

Simplifying:

$$y = x - \frac{g x^2}{u^2} \quad \text{(Equation 2)} $$

The ball just clears the wall at x = 4 m. Substituting x = 4 into Equation 2:

$$y = 4 - \frac{g (4)^2}{u^2} $$ $$ y = 4 - \frac{16g}{u^2} $$

From Equation 1, u² = 10g. Substituting this:

$$y = 4 - \frac{16g}{10g} $$

The g cancels out:

$$y = 4 - \frac{16}{10} $$

Simplifying the fraction:

$$\frac{16}{10} = 1.6 $$

Thus:

$$y = 4 - 1.6 = 2.4 \text{ m} $$

Therefore, the height of the wall is 2.4 m.

Comparing with the options:

A. 4.4 m

B. 2.4 m

C. 3.6 m

D. 1.6 m

Hence, the correct answer is Option B.