A particle is projected with velocity 2√(gh) so that it just clears two walls of equal height ‘h’ which are at a distance 2h from each other. The total horizontal range of the projectile is x√3h. The value of x is
Physics · Mixed Chapter
JEE Mains

Question

A particle is projected with velocity 2√(gh) so that it just clears two walls of equal height ‘h’ which are at a distance 2h from each other. The total horizontal range of the projectile is x√3h. The value of x is

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Launch speed u=2√(gh) ⇒u²=4gh. Let θ be the launch angle, and use the trajectory y=x tanθ−gx²/(2u²cos²θ). Step 2: At the two equal-height walls y=h. Putting X=x/h gives 1=X tanθ−X²/(8cos²θ). The two roots X₁,X₂ differ by 2 because walls are 2h apart. Step 3: From the quadratic-root separation, (X₂−X₁)² = (8sinθcosθ)²−32cos²θ=4. Hence 8cos²θ(2sin²θ−1)=1; this gives cos²θ=1/4 and θ=60°. Step 4: Total horizontal range R=(u²/g)sin2θ=4h sin120°=2√3 h. Comparing R=x√3 h yields x=2. Answer: 2.