Projectile Motion

For what angle of projection the horizontal range is equal to the max height:

• 45 degree

• tan $\theta$ =4

• tan $\theta$ =1/4

• 60 degree

Two bodies are projected at angles $\theta$  and $90-\theta$ . The ratio of their horizontal ranges will be 

• 1: tan $\theta$  

• tan $\theta$  :1

• 1:1

• sin $\theta$  :1

The vertical component of velocity at top most point is

• $u\cos\theta$

• u sin $\theta$

• zero

• u

A projectile is projected at an angle  $\theta$    so that its horizontal range is R, its max. height will be

• 1/4( R tan $\theta$ )

• 
  

  R/4

• 4 R tan $\theta$  

• 4 R

A ball is thrown vertically upwards with velocity 20 m/s by making an angle 30 degree with the vertical. Its horizontal component after time t = 1 s will be

• 10 m/s

• 10 $\sqrt{3}$ m/s

• 20 m/s

• None of the above

For a projectile the maximum horizontal range can be attained at an angle of projection (in degrees)

The formula for maximum horizontal range that can be achieved by a projectile is

• $u^2/2g$

• $u^2/4g$

• u/2g

• $u^2/g$

The value of acceleration in a projectile is (in m/s^2)

The acceleration along horizontal in a projectile is

The angle between the velocity and acceleration at top most point of a projectile is