A projectile is projected with initial velocity `(6hati+8hatj)` m/s.Minimum time after which its velocity becomes parallel to the line 4y-3x+2=0, is ?(`bar(g)=-10hatj`m/`s^(2)`)

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

A particle moving with initial velocity 4m/s decelerate at a constant rate of 2m/s^(2) .Time after which its velocity becomes zero is ?

A particle is projected with initial velocity of hati+2hatj . The eqaution of trajectory is (take g=10ms^(-2))

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

A particle is projected with a velocity vecv=8hati+6hatj m//s . The time after which it will starts moving perpendicular to its initial direction of motions is

What will be the equation for the path of a projectile if it is fired with initial velocity of (2hati+3hatj) . Here hatu and hatj are unit vectors along X-axis and Y-axis (g=10ms^(-2)) .

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

A projectile is given an initial velocity hati+2 hatj .The cartesin equation of its path is (g=10ms^(-2))

A cricketer hits a ball in a vertical x-y plane from the ground level with a velocity barV_0=(10sqrt(3i)+30hatj) m/s.Find the time in which velocity vector makes an angle of 30 ^(@) with horizontal x- axis (g=10 m//s^(2))

The velocity of a projectile at the initial point A is (2 hati +3 hatj) ms^(-1) . Its velocity (in ms^(-1)) at point B is