Home
Class 11
PHYSICS
The horizontal range of a projectile is ...

The horizontal range of a projectile is `4 sqrt(3)` times its maximum height. Its angle of projection will be

A

`45^(@)`

B

`60^(@)`

C

`90^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`R = 4sqrt(3) H`
`rArr (u^(2) sin 2theta)/g = 4sqrt(3) (u^(2) sin^(2) theta)/(2g)`
`rArr tan theta = 1/sqrt(3) rArr theta = 30^(@)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The horizontal range of a projectile is 2 sqrt(3) times its maximum height. Find the angle of projection.

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

The launching speed of a certain projectile is five times the speed it has at its maximum height.Its angles of projection is

The maximum horizontal range of a projectile is 400 m . The maximum value of height attained by it will be

Assertion: In projectile motion, when horizontal range is n times the maximum height, the angle of projection is given by tan theta=(4)/(n) Reason: In the case of horizontal projection the vertical increases with time.

The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

The maximum range of projectile is 2/sqrt3 times actual range. What isthe angle of projection for the actual range ?