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(D) The initial velocity components are:$u_x = u \cos 	heta = 40 \cos 30^{\circ} = 40 	imes \frac{\sqrt{3}}{2} = 20 \sqrt{3} \ m \ s^{-1}$$u_y = u \

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$20 \sqrt{13} \ m$

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(D) The initial velocity components are:$u_x = u \cos 	heta = 40 \cos 30^{\circ} = 40 	imes \frac{\sqrt{3}}{2} = 20 \sqrt{3} \ m \ s^{-1}$$u_y = u \sin 	heta = 40 \sin 30^{\circ} = 40 	imes \frac{1}{2} = 20 \ m \ s^{-1}$Horizontal displacement $(x)$ in time $t = 2.0 \ s$:$x = u_x 	imes t = (20 \sqrt{3}) 	imes 2 = 40 \sqrt{3} \ m$Vertical displacement $(y)$ in time $t = 2.0 \ s$:$y = u_y 	imes t - \frac{1}{2} g t^2 = (20 	imes 2) - \frac{1}{2} 	imes 10 	imes (2)^2 = 40 - 20 = 20 \ m$The net displacement $(S)$ is given by:$S = \sqrt{x^2 + y^2} = \sqrt{(40 \sqrt{3})^2 + (20)^2}$$S = \sqrt{1600 	imes 3 + 400} = \sqrt{4800 + 400} = \sqrt{5200}$$S = \sqrt{400 	imes 13} = 20 \sqrt{13} \ m$

$A$ projectile is launched with an initial speed of $40 \ m \ s^{-1}$ at an angle of $30^{\circ}$ above the ground. The projectile lands on a hillside $2.0 \ s$ later. The net displacement from where the projectile was launched to where it hits the target is (Take $g = 10 \ m \ s^{-2}$)

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