Home
Class 12
PHYSICS
The time of fight of an object projected...

The time of fight of an object projected with speed ` 20 ms^(-1)` at an angle ` 30^(@)` with the horizontal , is

A

1 s

B

4 s

C

2 s

D

6 s

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

The maximum height attained by a ball projected with speed 20 ms^(-1) at an angle 45^(@) with the horizontal is [take g = 10 ms^(-2) ]

A stone is projected with speed of 50 ms^(-1) at an angle of 60^(@) with the horizontal. The speed of the stone at highest point of trajectory is

An object is projected with a velocity of 30 ms^(-1) at an angle of 60^(@) with the horizontal. Determine the horizontal range covered by the object.

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

A particle is projected with a velocity of 20 ms^(-1) at an angle of 60^(@) to the horizontal. The particle hits the horizontal plane again during its journey. What will be the time of impact ?

Statement-1 : A man projects a stone with speed u at some angle. He again projects a stone with same speed such that time of flight now is different. The horizontal ranges in both the cases may be same. (Neglect air friction) Statement-2 : The horizontal range is same for two projectiles projected with same speed if one is projected at an angle q with the horizontal and other is projected at an angle (90^(@) theta) with the horizontal. (Neglect air friction)

A particle is projected with a velocity of 20sqrt(3) ms^(-1) at an angle of 60^(@) to the horizontal. At the moment, the direction of motion of the projectile is perpendicular to the direction of initial velocity of projection, its velocity is

A particle is projected with speed 20m s^(-1) at an angle 30^@ With horizontal. After how much time the angle between velocity and acceleration will be 90^@

A ball is projected from ground with a speed of 20ms^(-1) at an angle of 45^(@) with horizontal.There is a wall of 25m height at a distance of 10m from the projection point. The ball will hit the wall at a height of

What are the angles of projection of a projectile projected with velocity 30 ms^(-1) , so that the horizontal range is 45 m ? Take g = 10 ms^(-2) .