A particle is projected at an angle `theta` from ground with speed u (g = 10 m/`s^2` )

A particle is projected at an angle theta with an initial speed u .

A particle is projected at an angle 60^@ with horizontal with a speed v = 20 m//s . Taking g = 10m//s^2 . Find the time after which the speed of the particle remains half of its initial speed.

A particle is projected at an angle 60^@ with horizontal with a speed v = 20 m//s . Taking g = 10m//s^2 . Find the time after which the speed of the particle remains half of its initial speed.

A particle is projected at an angle 60^(@) with the horizontal with a speed 10m//s . Then, latus rectum is (Take g=10m//s^(2) )

A particle is projected from ground at an angle theta with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is -

A particle is projected from ground at an angle theta with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is -

A particle is projected at angle theta with horizontal from ground. The slop (m) of the trajectory of the particle varies with time (t) as

A particle is projected at angle theta with horizontal from ground. The slop (m) of the trajectory of the particle varies with time (t) as

A particle is projected from horizontal ground at angle 'theta' with speed 'u'. In same plane of motion a horizontal acceleration 'a' exists so that projected particle returns back to point of projection. Find time of flight.

A particle is projected from a point on the ground surface at different angles keeping the speed of projection as constant range obtained is plotted as function of the angle of projection with the horizontal 'O'. Find the range obtained if particle is projected at angle 30° with the horizontal : 2 g 10 m/s 250m R