JEE Advanced 2022 Paper 1 · Q08 · Projectile Motion

Question

A projectile is fired from horizontal ground with speed v and projection angle θ. When the acceleration due to gravity is g, the range of the projectile is d. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is g' = g/0.81, then the new range is d' = nd. The value of n is ___.

SolveKar AI · Accepted Answer

Original range d = v²sin 2θ/g; time to apex t₁ = v sin θ/g; horizontal distance covered to apex x₁ = v cosθ · t₁ = v² sin θ cosθ /g = d/2. Height at apex H = v² sin²θ/(2g). After apex, vertical velocity = 0, horizontal velocity = v cosθ, new gravity g' = g/0.81. Time to fall: t₂ = √(2H/g') = √(2·v² sin²θ/(2g)·0.81/g) = (v sinθ/g)·√0.81 = 0.9·v sinθ/g. Horizontal distance after apex x₂ = v cosθ · t₂ = 0.9 · v² sinθ cosθ/g = 0.9·d/2 = 0.45d. New total range d' = x₁ + x₂ = d/2 + 0.45d = 0.95d ⇒ n = 0.95.

JEE Advanced 2022 Paper 1 · Q08 · Projectile Motion

Solution