IIT Jee Problem: A particle is projected with a velocity 2 square Root ag. so that it just clears two walls of equal height 'a'.

A particle is projected with velocity 2 sqrt(gh) so that it just clears two walls of equal height h which are at a distance 2h from each other. Show that the time of passing between the walls is 2 sqrt(h//g) .

A particle is projected from the ground at t = 0 so that on its way it just clears two vertical walls of equal height on the ground. The particle was projected with initial velocity u and at angle theta with the horiozontal. If the particle passes just grazing top of the wall at time t = t_1 and t = t_2 , then calculate. (a) the height of the wall. (b) the time t_1 and t_2 in terms of height of the wall. Write the expression for calculating the range of this projectile and separation between the walls.

A piece of marble is projected from the earth's surface with a velocity of 50 m s^(-1) . 2 s later , it just clears a wall 5 m high. What is the angle of projection ?

A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

A particle is projected from the ground at an angle such that it just clears the top of a polar after t_1 time its path. It takes further t_2 time to reach the ground. What is the height of the pole ?

A particle is projected upwards with velocity 20m//s . Simultaneously another particle is projected with velocity 20(sqrt2) m//s at 45^@ . (g = 10m//s^2) (a) What is acceleration of first particle relative to the second? (b) What is initial velocity of first particle relative to the other? (c) What is distance between two particles after 2 s?

A perefectly elastic particle is projected with a velocity v in a vertical plane through the line of greatest slope of an inclined plane of elevation alpha . If after striking the plane , the particles rebounds vertically show that it will return to the point of projection at the end of time equal to (6v)/(gsqrt(1+8sin^(2) alpha))

A particle is projected with velocity u horizontally from the top of a smooth sphere of radius a so that it slides down the outside of the sphere. If the particle leaves the sphere when it has fallen a height (a)/(4) , the value of u is

A particle is projected with velocity v_(0) along x - axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=-x^(2) . The distance at which particle stops is -

A particle is projected from ground with velocity 20(sqrt2) m//s at 45^@ . At what time particle is at height 15 m from ground? (g = 10 m//s^2)