Home
Class 11
PHYSICS
A particle is projected with velocity 2(...

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 of 2h form each other. Show that the time of passing between the walls is `2 (sqrt (h/g)).` [Hint : First find velocity at height h. Treat it as initial velocity and 2h as the range. ]

Text Solution

Verified by Experts

The correct Answer is:
4
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KINEMATICS

    DC PANDEY ENGLISH|Exercise MATCH THE COLUMN|3 Videos

  • GRAVITATION

    DC PANDEY ENGLISH|Exercise (C) Chapter Exercises|45 Videos

  • KINEMATICS 1

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|15 Videos

Similar Questions

Explore conceptually related problems

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.

Knowledge Check

  • A ball of mass m is thrown vertically up with an initial velocity so as to reach a height h.The correct statement is :

    A
    Potential energy of the ball at the ground is mgh.
    B
    Kinetic energy of the ball at ground is zero
    C
    Kinetic energy of the ball at the highest point is mgh .
    D
    Potential enengy of the ball at the highest point is mgh.

    • Similar Questions

    Explore conceptually related problems

    A particle is projected from a point on the level ground and its height is h when at horizontal distances a and 2 a from its point of projection. Find the velocity of projection.

    a body is thrown horizontally with a velocity sqrt(2gh) from the top of a tower of height h. It strikes the level gound through the foot of the tower at a distance x from the tower. The value of x is :-

    A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

    A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

    A body projected horizontally with a velocity v from a height h has a range R .With what velocity a body is to be projected horizontally from a height h//2 to have the same range ?

    A particle is projected from ground with initial velocity u= 20(sqrt2)m//s at theta = 45^@ . Find (a) R,H and T, (b) velocity of particle after 1 s (c) velocity of particle at the time of collision with the ground (x-axis).

    A projectile is fired with velocity v_0 from a gun adjusted for a maximum range. It passes through two points P and Q whose heights above the horizontal are h each. Show that the separation of the two points is (v_0)/(g) sqrt(v_0^2 - 4 gh)) .

    Home
    Profile