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(A) Given: Initial speed $u = 50 \, m/s$,angle $	heta = 53^{\circ}$,time $t = 2 \, s$.The horizontal and vertical components of initial velocity are:

Vedclass · Accepted Answer

$60\sqrt{2}$

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(A) Given: Initial speed $u = 50 \, m/s$,angle $	heta = 53^{\circ}$,time $t = 2 \, s$.The horizontal and vertical components of initial velocity are:$u_x = u \cos 53^{\circ} = 50 	imes \frac{3}{5} = 30 \, m/s$$u_y = u \sin 53^{\circ} = 50 	imes \frac{4}{5} = 40 \, m/s$At any time $t$,the position coordinates $(x, y)$ of the projectile are given by:$x = u_x t = 30 	imes 2 = 60 \, m$$y = u_y t - \frac{1}{2} g t^2 = 40 	imes 2 - \frac{1}{2} 	imes 10 	imes (2)^2 = 80 - 20 = 60 \, m$The distance $d$ of the particle from the origin $(0, 0)$ at $t = 2 \, s$ is:$d = \sqrt{x^2 + y^2} = \sqrt{60^2 + 60^2} = \sqrt{2 	imes 60^2} = 60\sqrt{2} \, m$.

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