Question

A particle is projected perpendicularly to an inclined plane as shown in the figure. If the initial velocity of the particle is $u$, calculate how far from the point of projection does it hit the plane again (if the distance is measured along the plane) ?

• A. $\frac{2u^2}{g}$
• B. zero
• C. $\frac{2u^2}{g}\text{sin}\theta$
• D. $\frac{2u^2}{g}\text{tan}\theta \text{sec}\theta$

Answer 1

The correct option is D $\frac{2u^2}{g}\text{tan}\theta \text{sec}\theta$

Instead of taking the usual coordinate axis,
Let us change it such that y - axis is the direction of initial velocity
Now, components of acceleration due to gravity in the new coordinate system are $a_y=\text{g cos}\theta ,a_x=\text{g sin}\theta \text{and}{v}_y=u_f(u_x=0)$
All components are with respect to new coordinate axis
Now, taking displacement along inclined plane (x-axis)
$x=x_0+u_{x}t+\frac{1}{2}{a}_x{t}^2$
we have, $x_0=0,u_x=0,a_x=\text{g sin}\theta ,time=t$
$x=\frac{1}{2}\text{g sin}\theta t^2....\left(i\right)$
Take vertical motion (with respect to new axis) of projectile
(at highest point of trajectory)
$v_y=u_y-g\frac{t}{2}$
$v_y=0$ at highest point
$u_y=u$
$g=\text{g cos}\theta$
$0=u-\text{g cos}\theta \frac{t}{2}$
$\Rightarrow t=\frac{2u}{\text{g cos}\theta }$
substituting in equation (i),
$x=\frac{1}{2}\text{g sin}\theta \frac{4u^2}{g^{2}co{s}^2\theta }$
$=\frac{2u^2}{g}\frac{\text{sin}\theta }{\text{cos}\theta }\frac{1}{\text{cos}\theta }$
$=\frac{2u^2}{g}\text{tan}\theta \text{sec}\theta$