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JEE Advanced 2025 · Paper 2, Physics question 15

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 F⃗=−cv⃗\vec F = -c\vec v where the drag coefficient c=0.1c = 0.1 kg/s and v⃗\vec v is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking e=2.7e = 2.7, the horizontal distance of the wall from the point of projection (in m) is _____.

Answer and worked solution

Official answer: 167

Horizontal motion. Linear drag acts on each component separately: mv˙x=−cvxm\dot v_x=-cv_x, with cm=0.5\frac cm=0.5 s⁻¹. So
x(t)=v0cos⁡60∘c/m(1−e−ct/m)=270(1−e−1).x(t)=\frac{v_0\cos60^\circ}{c/m}\left(1-e^{-ct/m}\right)=270\left(1-e^{-1}\right).

At t = 2 s.
x=270(1−12.7)=170 m.x=270\left(1-\frac1{2.7}\right)=170\ \text{m}.

Topics:Projectile motionViscosity and Stokes' law
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