Question

A particle is projected up with a velocity of $v_0=10m/s$ at an angle of ${\theta }_0={60}^{\circ }$ with horizontal onto an inclined plane. The angle of inclination of the plane is ${30}^{\circ }$. the ratio of speeds of striking for upward and downward projection is :

Answer 1

Step $1$ : Draw rough diagram for the upward projection.


Step $2$ : Find time of flight.

Given, projection velocity, $v_0=10m/s$
Angle of inclination, $\alpha ={30}^{\circ }$
Angle of projection from horizontal, ${\theta }_0={60}^{\circ }$
So,
$\theta ={\theta }_0-\alpha$
$\theta ={60}^{\circ }-{30}^{\circ }={30}^{\circ }$
Time of flight for oblique projectile motion,
$T=\frac{2v_0\sin \theta }{g\cos \alpha }$
$T=\frac{2\times 10\times \sin {30}^{\circ }}{10\times \cos {30}^{\circ }}$
$T=\frac{2}{\sqrt{3}}s$

Step $3$ : Find speed of striking for upward projection.

From first equation of motion
$v=u+at$
$v_{xf}=v_0\cos {30}^{\circ }-g\sin {30}^{\circ }T$
$=10\frac{\sqrt{3}}{2}-\frac{10}{2}\frac{2}{\sqrt{3}}=\frac{5}{\sqrt{3}}m/s$
$v_{yf}=v_0\sin {30}^{\circ }-g\cos \alpha T$
$v_{yf}=\frac{10}{2}-10\times \frac{\sqrt{3}}{2}\times \frac{2}{\sqrt{3}}$
$=-5m/s$
$v_f=\sqrt{{\left(\frac{5}{\sqrt{3}}\right)}^2+5^2}=\frac{10}{\sqrt{3}}m/s$

Step $4$ : Draw rough diagram for the downward projection.


Step $5$ : Find speed of striking for downward projection.

Time of flight for downward projection will be same as upward projection, because $u_y$ is same in both cases.
From first equation of motion
$v=u+at$
$v_{xf}=v_0\cos {30}^{\circ }+g\sin {30}^{\circ }T$
$v_{xf}=\frac{25}{\sqrt{3}}$
$v_{yf}=v_0\sin {30}^{\circ }-g\cos \alpha T$
$v_{yf}=\frac{10}{2}-10\times \frac{\sqrt{3}}{2}\times \frac{2}{\sqrt{3}}$
$=-5m/s$
Then
$v_f=\sqrt{{\left(\frac{25}{\sqrt{3}}\right)}^2+5^2}=10\sqrt{\frac{7}{3}}$
So, ratio of speeds of striking for upward and downward projection $=1:\sqrt{7}$

Final answer: $\left(d\right)$