Odisha Board Solutions for Chapter: Projectile Motion, Exercise 1: MODEL QUESTIONS

A ball is projected from the top of a tower at an angle $60°$ with the vertical. What happens to the vertical component of velocity?

A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following remains constant?

A cricket ball is hit at $45°$ to the horizontal with a Kinetic energy $\mathrm{K}$. The kinetic energy at the highest point is

The angle of projection, for which the horizontal range and maximum height of a projectile are equal is

A particle is thrown horizontally from the top of a tower of height h with a velocity $\sqrt{2\mathrm{gh}}$. It strikes the ground at a distance from the foot of the tower equal to .

A particle is thrown horizontally from the top of a tower of height h with a velocity $\sqrt{2\mathrm{gh}}$. It strikes the ground at a distance from the foot of the tower equal to .

A particle is projected from the surface of earth vertically upward with a speed equal to half its escape velocity from earth. Taking earth to be a sphere of radius $\mathrm{R}$ calculate the height to which the particle rises above the surface of earth.

A particle is projected with velocity $2\sqrt{gh}$ so that it just clears two walls of equal height h which are at distance $2\mathrm{h}$ from each other. Prove that the time of flight between the two walls is $2\sqrt{\raisebox{1ex}{$\mathrm{h}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{g}$}\right.}$.