From a point on the ground at a distance 2 meters from the foot of a vertical wall, a ball is thrown at an angle of 45° which just clears the top of the wall and afterward strikes the ground at a distance 4 m on the other side. The height of the wall is

Projectile Motion Problem

Question: A ball is thrown from 2 meters away from a vertical wall at 45°. It just clears the wall and lands 4 meters beyond it. What is the wall's height?

Step-by-Step Solution

Given:

Total horizontal range = 2m (to wall) + 4m (beyond wall) = 6m

Key Formula (Projectile Range):

\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]

For θ = 45°, \(\sin(90°) = 1\):

\[ 6 = \frac{v_0^2}{9.8} \implies v_0^2 = 58.8 \implies v_0 \approx 7.67 \, \text{m/s} \]

Height Calculation at x = 2m:

\[ y = x \tan\theta - \frac{g x^2}{2v_0^2 \cos^2\theta} \]

Substitute values:

\[ y = 2(1) - \frac{9.8 \times 4}{2 \times 58.8 \times 0.5} = 2 - \frac{39.2}{58.8} \approx 1.33 \, \text{m} \]

Answer: The wall's height is \(\boxed{\frac{4}{3} \, \text{m}}\) (≈ 1.33 m).