A piece of marble is projected from the Earth | vedclass

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(D) The initial velocity is $u = 19.6 \sqrt{2} \ m/s$ at an angle $	heta = 45^{\circ}$.Horizontal component: $u_x = u \cos(45^{\circ}) = 19.6 \sqrt{2

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(D) The initial velocity is $u = 19.6 \sqrt{2} \ m/s$ at an angle $	heta = 45^{\circ}$.Horizontal component: $u_x = u \cos(45^{\circ}) = 19.6 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 19.6 \ m/s$.Vertical component: $u_y = u \sin(45^{\circ}) = 19.6 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 19.6 \ m/s$.After time $t = 2 \ s$,the horizontal velocity remains constant: $v_x = u_x = 19.6 \ m/s$.The vertical velocity is given by $v_y = u_y - gt = 19.6 - (9.8 	imes 2) = 19.6 - 19.6 = 0 \ m/s$.Since the vertical component of velocity is $0 \ m/s$,the velocity vector is purely horizontal.Therefore,the angle $\alpha$ with the horizontal is $0^{\circ}$.

$A$ piece of marble is projected from the Earth's surface with a velocity of $19.6 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$. $2 \ s$ later,its velocity makes an angle $\alpha$ with the horizontal,where $\alpha$ is $..........$ (in $^{\circ}$)

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