The initial velocity of a projectile at the point of projection is `3hat i+4hat j` `ms^(-1)` .Then the velocity of striking on the ground is

" The velocity of a projectile at point of projection is " 5hat i+4hat j " its velocity at the point of return to the same horizontal plane is "

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

The velocity of a projectile at the initial point A is (2hat(i)+3hat(j))m//s .Its velocity (in m//s) at point B is

At a height 0.4 m from the ground the velocity of a projectile in vector form is vec v = (6 hat i+ 2 hat j) ms ^-1 . The angle of projection is

The angular velocity of a rotating body is vec omega = 4 hat i + hat j - 2 hat k . The linear velocity of the body whose position vector 2hati + 3 hat j - 3 hat k is

A body with mass 5 kg is acted upon by a force vec(F) = (- 3 hat (i) + 4 hat (j)) N . If its initial velocity at t =0 is vec(v) = 6 hat(i) - 12 hat (j) ms^(-1) , the time at which it will just have a velocity along the y-axis is :

A projectile is given an initial velocity of (hat(i)+2hat(j)) The Cartesian equation of its path is (g = 10 ms^(-1)) ( Here , hati is the unit vector along horizontal and hatj is unit vector vertically upwards)

Velocity of a particle at some instant is v=(3hat i + 4hat j + 5hat k) m//s . Find speed of the particle at this instant.

A particle of mass 1 kg is projected with an initial velocity 10 ms^(-1) at an angle of projection 45^(@) with the horizontal. The average torque acting on the projectile and the time at which it strikes the ground about the point of projection in newton meter is

Writhe the projections of vec r=3hat i-4hat j+12hat k ion the coordinate axes.