A particle is projected with a velocity of 30,m / s , at an angle of θ_0=tan ^{-1}left(frac{3}{4}ri

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3-2.Motion in Plane

hard

A particle is projected with a velocity of $30\,m / s$, at an angle of $\theta_0=\tan ^{-1}\left(\frac{3}{4}\right)$ After $1\,s$, the particle is moving at an angle $\theta$ to the horizontal, where $\tan \theta$ will be equal to $\left(g=10\,m / s ^2\right)$

A

$1$

B

$2$

C

$\frac{1}{2}$

D

$\frac{1}{3}$

Solution

(d)

$u_x=u \cos \theta_0=20 \times \frac{4}{5}=24\,m / s$

and $\quad u_y=u \sin \theta_0=30 \times \frac{3}{5}=18\,m / s$

After $1 s , u_x$ will remain as it is $u_y$ will decreases by $10\,m / s$ or it will remain $8\,m / s$

$\therefore \quad \tan \theta=\frac{v_y}{v_x}=\frac{8}{24}=\frac{1}{3}$

Standard 11

Physics

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