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 ranges for upward and downward projection is :

• A. $1:1$
• B. $1:3$
• C. $1:2$
• D. $1:4$

Answer 1

The correct option is C $1:2$

Step $1$ : Draw rough diagram for the given situation.



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 range of the particle in the inclined plane for upward projection.

Range of the particle in the inclined plane,
$R=u_{x}T+\frac{1}{2}{a}_x{T}^2$
For upward projection, $a_x=-g\sin {30}^{\circ }$
$R_1=v_0\cos {30}^{\circ }T-\frac{1}{2}g\sin {30}^{\circ }{T}^2$
$=10\frac{\sqrt{3}}{2}\frac{2}{\sqrt{3}}-\frac{1}{2}\times 10\times \frac{1}{2}{\left(\frac{2}{\sqrt{3}}\right)}^2=\frac{20}{3}$

Step $4$ : Find range of the particle in the inclined plane for downward projection.

Range of the particle in the inclined plane,
$R=u_{x}T+\frac{1}{2}{a}_x{T}^2$
For downward projection, $a_x=g\sin {30}^{\circ }$
$R_2=v_0\cos {30}^{\circ }T+\frac{1}{2}gsin{30}^{\circ }{T}^2$
$=10\frac{\sqrt{3}}{2}\frac{2}{\sqrt{3}}+\frac{1}{2}\times 10\times \frac{1}{2}{\left(\frac{2}{\sqrt{3}}\right)}^2=\frac{40}{3}$

Step $5$ : Find the ratio of ranges.

The ratio of ranges for upward and downward projection,
$\frac{R_1}{R_2}=\frac{20}{3}\times \frac{3}{40}=\frac{1}{2}$

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