Question

A particle when projected in vertical plane moves along the smooth surface with initial velocity 20 m/s at an angle of ${60}^o$, so that its normal reaction on the surface remains zero throughout the motion. Then the slope of the tangent to the surface at height 5 m from the point of projection will be:

• A. ${30}^o$
• B. $ta{n}^{-1}\sqrt{2}$
• C. ${45}^o$
• D. $ta{n}^{-1}\left(2\right)$

Answer 1

The correct option is B

$ta{n}^{-1}\sqrt{2}$

For zero normal, particle always moves under the action of gravitational force i.e. projectile motion of particle.

$Now,v_y^2-u_y^2=2gh{v}_y^2-{\left(\frac{20\sqrt{3}}{2}\right)}^2=2(-10)\left(5\right)v_y=10\sqrt{2}m/stan\theta =\frac{v_y}{v_x}=\frac{10\sqrt{2}}{10}\theta =ta{n}^{-1}\left(\sqrt{2}\right)$