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Question

(C) Let the slope be the $x$-axis and the direction perpendicular to the slope be the $y$-axis.The initial velocity components are $u_x = v \sin 	het

Vedclass · Accepted Answer

$\frac{2v^2}{g} 	an 	heta \sec 	heta$

Vedclass · Answer

(C) Let the slope be the $x$-axis and the direction perpendicular to the slope be the $y$-axis.The initial velocity components are $u_x = v \sin 	heta$ and $u_y = v \cos 	heta$ is incorrect; rather,since the projectile is fired at a right angle to the slope,the initial velocity is $u_x = 0$ and $u_y = v$.The acceleration components are $a_x = -g \sin 	heta$ and $a_y = -g \cos 	heta$.For the projectile to land on the incline,the displacement in the $y$-direction must be zero: $y = u_y t + \frac{1}{2} a_y t^2 = 0$.$v t - \frac{1}{2} g \cos 	heta t^2 = 0 \implies t = \frac{2v}{g \cos 	heta}$.The range $R$ along the incline is the displacement in the $x$-direction at time $t$: $R = u_x t + \frac{1}{2} a_x t^2$.$R = 0(t) + \frac{1}{2} (-g \sin 	heta) \left( \frac{2v}{g \cos 	heta} ight)^2$.$R = -\frac{1}{2} g \sin 	heta \left( \frac{4v^2}{g^2 \cos^2 	heta} ight) = -\frac{2v^2 \sin 	heta}{g \cos^2 	heta} = -\frac{2v^2}{g} 	an 	heta \sec 	heta$.Taking the magnitude,$R = \frac{2v^2}{g} 	an 	heta \sec 	heta$.

$A$ projectile is fired with a velocity $v$ at a right angle to a slope which is inclined at an angle $\theta$ with the horizontal. The expression for the range $R$ along the incline is

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