Home
Class 11
PHYSICS
A particle is projected at an angle of e...

A particle is projected at an angle of elevation `alpha` and after t second it appears to have an elevation of `beta` as seen from the point of projection. Find the initial velocity of projection.

A

`("gt")/(2sin(alpha-beta))`

B

`("gt"cosbeta)/(2sin(alpha-beta))`

C

`(sin(alpha-beta))/(2"gt")`

D

`(2sin(alpha-beta))/("gt"cosbeta)`

Text Solution

Verified by Experts

The correct Answer is:
B
b. `y/x = tan beta`
`rArr (usin alphat - 1/2 "gt"^2)/(u cos alphat) = tanbeta`
or `usin alpha-tan betaucosalpha = ("gt")/2 or u = (gtcosbeta)/(2sin(alpha-beta))`.
.
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS KINEMATICS

    CENGAGE PHYSICS|Exercise Multiple Correct Answer Type|26 Videos

  • MISCELLANEOUS KINEMATICS

    CENGAGE PHYSICS|Exercise Linked Comprehension Type|35 Videos

  • LINEAR AND ANGULAR SIMPLE HARMONIC MOTION

    CENGAGE PHYSICS|Exercise Multiple Correct Answer Type|9 Videos

  • MISCELLANEOUS VOLUME 2

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|10 Videos

Similar Questions

Explore conceptually related problems

A particle I projected at an angle of elevation alpha after time t it makes an angle beta with horizontal. The velocity of projection is

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

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.

If an object is thrown at an angle of 60^(@) with horizontal, find elevation angle of the object at its highest point as seen from the point of projection.

A ball is projected from ground in such a way that after 10 seconds of projection it lands on ground 500 m away from the point of projection. Find out :- (i) angle of projection (ii) velocity of projection (iii) Velocity of ball after 5 seconds

A particle is projected with velocity v at an angle theta aith horizontal. The average angle velocity of the particle from the point of projection to impact equals

A particle is projected from the ground with an initial speed v at an angle theta with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is [EAM 2013]

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.