JEE Advanced 2025 — Paper 2 — Question 15

$\vec{F}_{\text {net }}=m \frac{d \vec{v}}{d t}$

$\mathrm{mg}+\overrightarrow{\mathrm{F}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}$

$\mathrm{mg}-\mathrm{C} \overrightarrow{\mathrm{v}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}$

Horizontal direction$-C v_{x}=\frac{m d v_{x}}{d t}$

$-\frac{\mathrm{C}}{\mathrm{m}} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{\mathrm{v}_{0 \mathrm{x}}}^{\mathrm{v}_{\mathrm{x}}} \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{x}}}$

$-\frac{\mathrm{t}}{2}=\ln \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{v}_{0 \mathrm{x}}}$

$\frac{d x}{d t}=v_{x}=v_{0 x} e^{-t / 2}$

$\int_{0}^{S_{x}} d x=V_{0 x} \int_{0}^{t} e^{-t / 2} d t$

$S_{x}=2 v_{0 x}\left(1-e^{-t / 2}\right)$ at $\mathrm{t}=2 \mathrm{sec}$

$\mathrm{S}_{\mathrm{x}}=2 \times 270 \times \cos 60^{\circ}\left[1-\frac{1}{\mathrm{e}}\right]$

$\mathrm{S}_{\mathrm{x}}=270\left(1-\frac{1}{2.7}\right)$

$=\frac{270}{2.7} \times(1.7)$

$=170 \mathrm{~m}$

$\mathrm{S}_{\mathrm{x}}=170 \mathrm{~m}$