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A particle of mass m is projected with a speed u from the ground at an angle θ = π/3 w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity u i. The horizontal distance covered by the combined mass before reaching the ground is

1. 3√3 u² / 8g
2. 2√2 u² / g
3. 5u² / 8g
4. 3√2 u² / 4g

Correct answer: 3√3 u² / 8g

Solution

At apex vx=u/2, vy=0, height H=u^2 sin^2(60)/(2g)=3u^2/(8g). Inelastic collision with mass m moving at u gives combined vx=(u/2+u)/2=3u/4. Fall time t=sqrt(2H/g)=u*sqrt(3)/(2g), so horizontal distance = (3u/4)*t = 3*sqrt(3)*u^2/(8g).

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