The surface of a hill is inclined at an angle | vedclass

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Question

(C) Let the angle of projection with the horizontal be $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The angle of inclination of the hill is $\alph

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Let the angle of projection with the horizontal be $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The angle of inclination of the hill is $\alpha = 30^{\circ}$.The range $R$ along the inclined plane is given by the formula:$R = \frac{2 u^2 \cos 	heta \sin(	heta - \alpha)}{g \cos^2 \alpha}$However, using the components along and perpendicular to the incline:Initial velocity component along the incline: $u_x = u \cos(	heta + \alpha) = 10 \cos(30^{\circ} + 30^{\circ}) = 10 \cos 60^{\circ} = 5 	ext{ m/s}$.Initial velocity component perpendicular to the incline: $u_y = u \sin(	heta + \alpha) = 10 \sin(60^{\circ}) = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3} 	ext{ m/s}$.Acceleration components: $a_x = g \sin 30^{\circ} = 5 	ext{ m/s}^2$ and $a_y = -g \cos 30^{\circ} = -5\sqrt{3} 	ext{ m/s}^2$.Time of flight $T$ is when displacement perpendicular to the incline is zero:$0 = u_y T + \frac{1}{2} a_y T^2 \implies T = \frac{-2 u_y}{a_y} = \frac{-2(5\sqrt{3})}{-5\sqrt{3}} = 2 	ext{ s}$.Distance $AB = u_x T + \frac{1}{2} a_x T^2 = 5(2) + \frac{1}{2}(5)(2^2) = 10 + 10 = 20 	ext{ m}$.

Similar Questions

$A$ projectile is launched with a speed of $10\,m/s$ at an angle of $60^{\circ}$ with the horizontal from a sloping surface of inclination $30^{\circ}$. The range $R$ is $........\,m$ (Take $g=10\,m/s^2$).

$A$ projectile is projected with speed $u$ at an angle of $60^o$ with the horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally,the range on the inclined plane will be:

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