AIIMS 2018 Physics Range of Projectile MCQ Question

To determine how far beyond the wall the body will strike the ground, we first calculate its horizontal range. The horizontal component of the initial velocity (v₀ₓ) is given by v₀ * cos(θ) = 50 m/s * cos(30°) = 43.3 m/s. In 3 seconds, the horizontal distance traveled is 43.3 m/s * 3 s = 129.9 m.

Next, we find the vertical position of the body at 3 seconds using the vertical component (v₀ᵧ = 50 m/s * sin(30°) = 25 m/s) and the equation of motion: y = v₀ᵧ * t - (1/2) * g * t² = 25 m/s * 3 s - 0.5 * 9.8 m/s² * (3 s)² = 75 m - 44.1 m = 30.9 m.

The body will strike the ground at a horizontal distance of 129.9 m from the launch point. Since the wall is at 43.3 m, the distance beyond the wall is 129.9 m - 43.3 m = 86.6 m, confirming option A. Options B, C, and D are incorrect as they do not account for the correct horizontal distance traveled after the wall.