Home
Class 11
PHYSICS
A particle is projected form ground with...

A particle is projected form ground with velocity u ar angle `theta` from horizontal. Match the following two columns.

Text Solution

Verified by Experts

The correct Answer is:
`A rarr q,B rarr s, C rarr p, D rarr s`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOTION

    DC PANDEY ENGLISH|Exercise C. Medical entrances gallery|1 Videos

  • MOTION

    DC PANDEY ENGLISH|Exercise Medical entrances gallery|19 Videos

  • MOTION

    DC PANDEY ENGLISH|Exercise Medical entrance|12 Videos

  • MEASUREMENT AND ERRORS

    DC PANDEY ENGLISH|Exercise Subjective|19 Videos

  • MOTION IN A PLANE

    DC PANDEY ENGLISH|Exercise (C )Medical entrances gallery|32 Videos

Similar Questions

Explore conceptually related problems

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal range, maximum height and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

A particle is projected from the ground with velocity u making an angle theta with the horizontal. At half of its maximum heights,

A particle is projected from ground with speed u and at an angle theta with horizontal. If at maximum height from ground, the speed of particle is 1//2 times of its initial velocity of projection, then find its maximum height attained.

A particle of mass m is projected form ground with velocity u making angle theta with the vertical. The de Broglie wavelength of the particle at the highest point is

A particle is projected from the ground with speed u at angle 60^@ from horizontal. It collides with a second particle of same mass moving with horizontal speed u in same direction at highest point of its trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision when they reached at ground

A particle is projected from a point P with a velocity v at an angle theta with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

A particle is projected from a point P with a velocity v at an angle theta with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

A particle is projected with velocity u at angle theta with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

A particle is projected with velocity u at angle theta with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

A particle is projected with velocity 'u' makes an angle theta w.r.t. horizontal. Now it breaks in two identical parts at highest point of trajectory. If one part is retrace its path, then velocity of other part is -

Home
Profile