A body projected along a frictionless incline | vedclass

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(C) The effective acceleration along the direction $BA$ is $a = g \sin 30^{\circ}$.$a = 10 	imes \frac{1}{2} = 5\, m/s^2$.Since the body stops after

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$20 \sqrt{3}\, m$

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(C) The effective acceleration along the direction $BA$ is $a = g \sin 30^{\circ}$.$a = 10 	imes \frac{1}{2} = 5\, m/s^2$.Since the body stops after covering a distance $s = 40\, m$ along the plane $AB$,we use the equation of motion $v^2 = u^2 + 2as$.Here,$v = 0$ (final velocity),$a = -5\, m/s^2$ (retardation),and $s = 40\, m$.$0 = u^2 - 2 	imes 5 	imes 40 \Rightarrow u^2 = 400 \Rightarrow u = 20\, m/s$.Now,if the particle is projected with this speed $u = 20\, m/s$ at an angle $	heta = 30^{\circ}$ with the horizontal,the range $R$ is given by:$R = \frac{u^2 \sin 2	heta}{g} = \frac{(20)^2 \sin(2 	imes 30^{\circ})}{10}$.$R = \frac{400 	imes \sin 60^{\circ}}{10} = 40 	imes \frac{\sqrt{3}}{2} = 20 \sqrt{3}\, m$.

$A$ body projected along a frictionless inclined plane of angle of inclination $30^{\circ}$ covers a distance of $40\, m$ along the plane. If the body is projected with the same speed at an angle of $30^{\circ}$ with the ground,it will have a range of: (Take $g = 10\, m/s^2$)

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