A body is projected with a velocity V = 3`hati`+ 4`hatj`+ 12`hatk` units. Assume xy plane is horizontal (g=10m/s). The range and maximum height are?

A body is projected with a initial velocity of (8hati+6hatj) m s^(-1) . The horizontal range is (g = 10 m s^(-2))

A body is projected with a velocity vecv =(3hati +4hatj) ms^(-1) The maximum height attained by the body is: (g=10 ms^(-2))

A body is projected with velocity u so that its horizontal range is twice the maximum height.The range is

A body is projected from ground with velocity u making an angle theta with horizontal. Horizontal range of body is four time to maximum height attained by body. Find out (i) angle of projection (ii) Range and maximum height in terms of u (iii) ratio of kinetic energy to potential energy of body at maximum height

A body is projected with a velocity of 60 m/s at 30^(@) to horizontal (assume positive x-axis and positive y axis as horizontal and vertical directions, respectively, and take g=10m//s^(2) ).

A body is prodected with initial velocity ( 5hati + 12hati) m//s from origin . Gravity acts in negative y direction. The horizontal rang is (Take g = 10 m//s^(2) )

A particle is projected from ground with velocity 3hati + 4hati m/s. Find range of the projectile :-

A body is projected from the ground with a velocity v = (3hati +10 hatj)ms^(-1) . The maximum height attained and the range of the body respectively are (given g = 10 ms^(-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 body is projected with velocity u such that in horizontal range and maximum vertical heights are samek.The maximum height is