JEE Advanced 2025 — Paper 2 — PHY-III solution

JEE Advanced2025Paper 2PHY-III

A projectile of mass 200 g is launched in a viscous medium at an angle 60 $^{\circ}$ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $F = - c v$ where the drag coefficient $c = 0.1$ kg/s and $v$ is the instantaneous velocity of the projectile. 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 ____

Your answer

Solution

The horizontal equation of motion has only the drag term:

$m \frac{d v _{x}}{d t} = - c v_{x} \Rightarrow v_{x} (t) = v_{0 x} e^{- (c / m) t}$

With $m = 0.2$ kg and $c = 0.1$ kg/s, the time constant is $m / c = 2$ s, so $v_{x} (t) = v_{0 x} e^{- t /2}$ .

The horizontal displacement up to time $t$ :

$S_{x} = \int_{0}^{t} v_{0 x} e^{- t^{'} /2} d t^{'} = 2 v_{0 x} (1 - e^{- t /2})$

At $t = 2$ s and $v_{0 x} = 270 cos 6 0^{\circ} = 135$ m/s:

$S_{x} = 2 \times 135 \times (1 - \frac{1}{e}) = 270 \times (1 - \frac{1}{2.7}) = 270 \times \frac{1.7}{2.7} = 170$ m.

$∴ S_{x} = 170.00$ m.

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