Question

A particle is projected from the bottom of an inclined plane of inclination ${30}^{\circ }$ with velocity of 40 m/s at an angle of ${60}^{\circ }$ with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take $(g=10m/s^2)$.

Answer 1

Given : $u=40m/s$

Velocity vector $u$ makes an angle ${30}^o$ with the inclined plane.

From figure $u_x=ucos\left({30}^o\right)$ m/s and $u_y=usin\left({30}^o\right)$ m/s

Also $a_x=-gsin\left({30}^o\right)$ $m/s^2$ and $a_y=-gcos\left({30}^o\right)$ $m/s^2$

Let the velocity is parallel to the inclines plane i.e $V_y=0$ at time $=t$

y direction : $V_y=u_y+a_{y}t$

$\therefore$ $0=usin30-\left(gcos30\right)t$

$⟹$ $t=\frac{utan{30}^o}{g}=\frac{40}{10\times \sqrt{3}}=\frac{4}{\sqrt{3}}$ s

x direction : $V_x=u_x+a_{x}t$

$V_x=ucos30-\left(gsin30\right)t$

$\therefore$ $40\times \frac{\sqrt{3}}{2}-10\times 0.5\times \frac{4}{\sqrt{3}}=\frac{40}{\sqrt{3}}$ m/s

Thus the speed of the particle $V=V_x=\frac{40}{\sqrt{3}}$ m/s