JEE Advanced 2025 Paper 2 · Q15 · Projectile Motion
A projectile of mass $200$ g is launched in a viscous medium at an angle $60°$ with the horizontal, with an initial velocity of $270$ m/s. It experiences a viscous drag force $\vec{F} = -c\vec{v}$ where the drag coefficient $c = 0.1$ kg/s and $\vec{v}$ is the instantaneous velocity. The projectile hits a vertical wall after $2$ s. Taking $e = 2.7$, the horizontal distance of the wall from the point of projection (in m) is __________.
Reveal answer + step-by-step solution
Correct answer:170
Solution
Horizontal equation: $m\,dv_x/dt = -c v_x \to v_x(t) = v_0 \cos(60°)\,e^{-(c/m) t} = 135\,e^{-(0.1/0.2) t} = 135\,e^{-t/2}$. $x(t) = \int_0^t v_x\,d\tau = 135 \times 2(1 - e^{-t/2}) = 270(1 - e^{-t/2})$. At $t = 2$ s: $e^{-1} = 1/e = 1/2.7$. $x = 270(1 - 1/2.7) = 270(1.7/2.7) = 270 \times 17/27 = 170$ m.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →