A particle has an initial velocity `(12hati+10hatj) ms^(-1)` and an acceleration of `(0.5hati+0.6hatj)ms^(-2)`. Its speed after 20s is

A particle has an initial velocity (6hati+8hatj) ms^(-1) and an acceleration of (0.8hati+0.6hatj)ms^(-2) . Its speed after 10s is

A particle has an initial velocity (3hati+2hatj) ms^(-1) and an acceleration of (0.2hati+0.3hatj)ms^(-2) . Its speed after 10s is

A particle has initial velocity (5hati+7hatj) and acceleration (0.7hati+0.5hatj) . The magnitude of velocity after 10 second will be

A particle has velocity (6t^3hati+8t^2hatj) ms^(-1) find the velocity of the particle

A cricket ball of mass 150 g has an initial velocity mu=(3hati+4hatj) ms^(-1) and a final velocity v = - (3hat(i) + 4hat(j)) ms^(-1) , after being hit. The change in momentum (final momentum­initial momentum) is (in kgms^(-1) )

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

In a reference frame a man A is moving with velocity (3hati-4hatj)ms^(-1) and another man B is moving with velocity (hati+hatj)ms^(-1) relative to A. Find the actual velocity of B.

A light body is projected with a velocity (10hat(i)+20 hat(j)+20 hat(k)) ms^(-1) . Wind blows along X-axis with an acceleration of 2.5 ms^(2) . If Y-axis is vertical then the speed of particle after 2 second will be (g=10 ms^(2))

A projectile is thrown with an initial velocity of (a hati +b hatj) ms^(-1) . If the range of the projectile is twice the maximum height reached by it, then

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