Question

A particle is fired horizontally from an inclined plane of inclination ${30}^0$ with horizontal with speed $50m{s}^{-1}.$ If $g=10m{s}^{-2},$ the range measured along the incline

• A. 500 m
• B. $\frac{1000}{3}m$
• C. $200\sqrt{2}m$
• D. $100\sqrt{3}m$

Answer 1

The correct option is B $\frac{1000}{3}m$

$a_x=gsin{30}^0=\frac{g}{2},u_x=ucos{30}^0=\frac{\sqrt{3}u}{2}=25\sqrt{3}m/s{a}_y=-gcos{30}^0=\frac{-\sqrt{3}g}{2},u_y=usin{30}^0=\frac{u}{2}=25m/sy=0\Rightarrow utsin{30}^0+\frac{-\sqrt{3}g{t}^2}{4}=0\Rightarrow T=\frac{4usin{30}^0}{\sqrt{3}g}=\frac{10}{\sqrt{3}}sR=uTcos{30}^0+\frac{g{T}^2}{4}=25\sqrt{3}\times \frac{10}{\sqrt{3}}+\frac{10\times 100}{12}=333.33$