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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Advanced 2025 · Paper 2 · Physics · Motion in Two Dimensions

A projectile of mass 200 g is launched in a viscous medium at an...

A projectile of mass 200 g is launched in a viscous medium at an angle 60∘60^{\circ} with the horizontal, with an initial velocity of 270 m/s270 \mathrm{~m} / \mathrm{s}. It experiences a viscous drag force F⃗=−cv⃗\vec{F}=-c \vec{v} where the drag coefficient c=0.1 kg/sc=0.1 \mathrm{~kg} / \mathrm{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

Correct answer

170

Step-by-step solution

\overrightarrow{\mathrm{F}_{\mathrm{net}}=\mathrm{m} \frac{\mathrm{ d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} \mathrm{mg}+\overrightarrow{\mathrm{F}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} \mathrm{mg}-\mathrm{C} \overrightarrow{\mathrm{v}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} Horizontal direction −Cvx=mdvxdt-\mathrm{Cv}_{\mathrm{x}}=\frac{\mathrm{mdv}_{\mathrm{x}}}{\mathrm{dt}}−Cm∫0tdt=∫v0xvxdvxvx−t2=ℓnvxv0xdxdt=vx=v0xe−t/2∫0Sxdx=v0x∫0te−t/2dtSx=2v0x(1−e−t/2) at t=2sec⁡\begin{aligned} & -\frac{C}{m} \int_0^t d t=\int_{v_{0 x}}^{v_x} \frac{d v_x}{v_x} \\ & -\frac{t}{2}=\ell n \frac{v_x}{v_{0 x}} \\ & \frac{d x}{d t}=v_x=v_{0 x} e^{-t / 2} \\ & \int_0^{S_x} d x=v_{0 x} \int_0^t e^{-t / 2} d t \\ & S_x=2 v_{0 x}\left(1-e^{-t / 2}\right) \\ & \text{ at } t=2 \sec \end{aligned}Sx=2×270×cos⁡60∘[1−1e]Sx=270(1−12.7)=2702.7×(1.7)=170mSx=170m\begin{aligned} & \mathrm{S}_{\mathrm{x}}=2 \times 270 \times \cos 60^{\circ}\left[1-\frac{1}{\mathrm{e}}\right] \\ & \mathrm{S}_{\mathrm{x}}=270\left(1-\frac{1}{2.7}\right) \\ & =\frac{270}{2.7} \times(1.7) \\ & =170 \mathrm{ m} \\ & \mathrm{ S}_{\mathrm{x}}=170 \mathrm{ m}\end{aligned}$$
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