Home
Class 11
PHYSICS
The launching speed of a certain project...

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

A

`theta = cos^(-1) (0.2)`

B

`theta = sin^(-1) (0.2)`

C

`theta = tan^(-1) (0.2)`

D

`theta = 0^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`u =5u cos theta rArr cos theta =(1)/(5)`
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

The speed of a projectile when it is at its greatest height is sqrt(2//5) times its speed at half the maximum height. The angle of projection is

The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

" For a projectile the ratio of the square of the time of flight to its maximum height is "

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is .

An object is projected with speed u and range of the projectile is found to be double of the maximum height attained by it . Range of the projectile is .

An object is projected with speed u and range of the projectile is found to be double of the maximum height attained by it . Range of the projectile is .