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JEE ADVANCED 2025
Paper-2 2025
Question
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 $240 \mathrm{~m} / \mathrm{s}$. It experiences a viscous drag force $\vec{F}=-c \vec{v}$ where the drag coefficient $\mathrm{c}=0.1 \mathrm{~kg} / \mathrm{s}$ and $\vec{v}$ 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 $\_\_\_\_$
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Solution

$$ \begin{aligned} & v=w_i\left(w_i\right) \\ & F=0.1 \mathrm{kw}\left(v^2+v_{,} \hat{i}\right)-10 \mathrm{~m} \\ & d a-\left(\frac{\bar{w} \bar{i}+w_j \bar{j}}{z}\right)-10 \bar{j} \\ & d=-\frac{w_k}{z} i-\left(\frac{w_p+20}{z}\right) \bar{p} \\ & \frac{d w_n}{d r} \frac{-w_n}{2} \\ & \int_{n=1}^n \frac{d v_n}{q_n}-v_n \frac{d}{2} \\ & \ln \left(\frac{v}{135}\right)=-\frac{r}{2} \\ & p=135- \\ & \frac{d x}{d t}=13 e^{-} \\ & \int d x=13 e^{t^2} e^{-x t} \\ & =13.5-21[-1+\mathrm{c}] \\ & =270\left(1-\frac{1}{2}\right) \\ & =270\left(1-\frac{1}{27}\right) \\ & =270 \times \frac{1.7}{27} \\ & =170 \end{aligned} $$
Question Tags
JEE Advance
Physics
Easy
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