Question

A particle when projected in vertical plane moves along smooth surface with initial velocity $20m{s}^{-1}$ at an angle of ${60}^{\circ },$ so that its normal reaction on the surface remains zero throughout the motion. Then the slope of the surface at height $5m$ from the point of projection will be:

• A. $\sqrt{3}$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (D)

None of these

Horizontal velocity always remains same.

We use $v^2-u^2=2as$ to calculate the final vertical velocity of the block.

On travelling 5m, the verticle velocity of the particle will be $10\sqrt{2}m/s$

Therefore, slope will be $=\frac{verticalvelocity}{horizontalvelocity}=\frac{10\sqrt{2}}{10}$

Slope $=\sqrt{2}$