A projectile is projected from top of a wall 20m height with a speed of 10m/s horizontally. Find range `(g = 10m/s^(2))`

A projectile is projected with the initial velocity (6i + 8j) m//s . The horizontal range is (g = 10 m//s^(2))

A projectile is projected from horizontal ground with velocity 18km/hr at an angle of 60^(@) from horizontal. Find angular speed (in rad/s) as observed from the point of projection at the time of landing.[ g=10m/ s^(2) ]

A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

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) )

From the ground, a projectile is fired at an angle of 60 degree to the horizontal with a speed of 20" m s"^(-1) The horizontal range of the projectile is

A particle of mass 50 g is projected horizontally from the top of a tower of height 50 m with a velocity 20m//s . If g=10m//s^(2) , then find the :-

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 projectile is projected with a speed = 20 m/s from the floor of a 5 m high room as shown. Find the maximum horizontal range of the projectile and the corresponding angle of projection theta .

A ball is projected vertically up with speed 20 m/s. Take g=10m//s^(2)

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