A particle is projected at an angle of `45^(@)` with a velocity of `9.8 ms^(-1)`. The horizontal range will be (Take, `g = 9.8 ms^(-2))`

A particle is projected at an angle of 45^@ with a velocity 9.8ms^-1 . What is horizontal range.

A ball is projected at an angle 30° with the horizontal with the velocity 49 ms^(-1) . The horizontal range is

A ball is projected at an angle 30° with the horizontal with the velocity 49 ms^(-1) . The horizontal range is

A particle is projected at an angle theta =30^@ with the horizontal, with a velocity of 10ms^(-1) . Then

A particle is projected at an angle theta =30^@ with the horizontal, with a velocity of 10ms^(-1) . Then

A particle is projected from horizontal making an angle of 53^(@) with initial velocity 100 ms^(-1) . The time taken by the particle to make angle 45^(@) from horizontal is

A particle is projected from horizontal making an angle of 53^(@) with initial velocity 100 ms^(-1) . The time taken by the particle to make angle 45^(@) from horizontal is

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) .

A body is projected at angle 30° to horizontal with a velocity 50 ms^(-1) .In the above problem range of body is m. [g = 10 ms^(-2) ]