Question

A particle is projected from point P with velocity 5√2 ms^-1 perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. The time of the flight of the particle is

• A. 1 sec.
• B. √2 sec.
• C. 2 √2 sec
• D. 2 sec

Answer 1

Answer: (A)

1 sec.

Answer 2

t = $\frac{u}{g\sin\theta} = \frac{5\sqrt{2} \times \sqrt{2}}{10}$ = 1 sec. Hence (1) is correct.