A bullet is fired at time t=0 with velocity 2 | vedclass

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(B) The horizontal component of velocity is $u_x = u \cos 	heta = 20 \cos 30^{\circ} = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \ m/s$.The vertical

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$\frac{19}{20 \sqrt{3}}$

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(B) The horizontal component of velocity is $u_x = u \cos 	heta = 20 \cos 30^{\circ} = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \ m/s$.The vertical component of velocity is $u_y = u \sin 	heta = 20 \sin 30^{\circ} = 20 	imes \frac{1}{2} = 10 \ m/s$.After time $t = 0.1 \ s$, the horizontal displacement is $x = u_x t = 10\sqrt{3} 	imes 0.1 = \sqrt{3} \ m$.The vertical displacement is $y = u_y t - \frac{1}{2}gt^2 = 10 	imes 0.1 - \frac{1}{2} 	imes 10 	imes (0.1)^2 = 1 - 0.05 = 0.95 \ m = \frac{19}{20} \ m$.The angle $	heta$ between the displacement vector and the horizontal is given by $	an 	heta = \frac{y}{x}$.Substituting the values, $	an 	heta = \frac{19/20}{\sqrt{3}} = \frac{19}{20\sqrt{3}}$.Thus, $	heta = 	an^{-1}\left(\frac{19}{20\sqrt{3}}ight)$. The question asks for the value of $	an 	heta$ (implied by options). Therefore, the correct value is $\frac{19}{20\sqrt{3}}$.

$A$ bullet is fired at time $t=0$ with velocity $20 \ m/s$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time $0.1 \ s$ is (Assume $g=10 \ m/s^2$)

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