What is the range of a projectile thrown with velocity `98 m//s` with angle `30^(@)` from horizontal ?

The angle which the velocity vector of a projectile thrown with a velocity v at an. angle theta to the horizontal Will make with the horizontal after time t of its being thrown up is

The angle of which the velocity vector of a projectile thrown with a velocity u at an angle theta to the horizontal will take with the horizontal after time t of its being thrown up is

The angle of which the velocity vector of a projectile thrown with a velocity u at an angle theta to the horizontal will take with the horizontal after time t of its being thrown up is

Figure shows a projectile thrown with speed u = 20 m/s at an angle 30^(@) with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is

Figure shows a projectile thrown with speed u = 20 m/s at an angle 30^(@) with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is

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

A particle is thrown at time t=0 with a velocity of 10 m/s at an angle of 60^(@) with the horizontal from a point on an incline plane, making an angle of 30^(@) with the horizontal. The time when the velocity of the projectile becomes parallel to the incline is :

What are two angles of projection of a projectile projected with velocity 30 m/s , so that the horizontal range is 45m . Take g=10 m/s^2 .