A ball is projected from the top of a tower at an angle `60^(@)` with the vertical. What happens to the vertical component of its velocity?

A body is projected horizontally from the top of a tower with a velocity of 20 m/s. After what time the vertical component of velocity is four times the horizontal component of velocity ? (g = 10 m//s^(2) )

A body is projected horizontally from the top of a tower with a velocity of 20 m/s. After what time the vertical component of velocity is four times the horizontal component of velocity ? (g = 10 m//s^(2) )

A body is projected horizontally from the top of a tower with a velocity of 20m/s .After what time the vertical component of velocity is four times the horizontal component of velocity (g=10m/s^(2))

A body is projected with an initial velocity of 58.8 m/s at angle 60° with the vertical. The vertical component of velocity after 2 sec is

A body is projected with an initial velocity of 58.8 m/s at angle 60° with the vertical. The vertical component of velocity after 2 sec is

A body is projected horizontally from the top of a tower with initial velocity 18 m s^-1 . It hits the ground at angle 45^@ . What is the vertical component of velocity when it strikes the ground ?