Question

A particle is perpendicular to an inclined plane as shown.If the initial velocity of the particle is u, calc how far from the point of projection does it hit plane again if the distance is measured along the plane?

• A. $2u^2/g$
• B. 0
• C. $\left(2u^2/g\right)sin\theta$
• D. $\left(2u^2/g\right)tan\theta sec\theta$

Answer 1

The correct option is D $\left(2u^2/g\right)tan\theta sec\theta$

time of flight

$S_y=0$

$S_y=u_{y}T+\frac{1}{2}{a}_y{t}^2$

$0=uT-\frac{1}{2}gcos\theta T^2$

$T=\frac{2u}{gcos\theta }$

range

$R=u_{x}T+\frac{1}{2}{a}_x{T}^2$

$=0+\frac{1}{2}gsin\theta \frac{4u^2}{g^2{cos}^2\theta }$

$=\frac{2u^{2}sin\theta }{gcos\theta cos\theta }$

=$\frac{2u^2}{g}tan\theta sec\theta$