A ball of 1.25 kg is traveling at 19 m/s towards the north. There is a force acting on it of 23 N in a direction 30 degrees north of east. What is the velocity of the ball after 12 s.
The ball's initial velocity is 19 m/s towards the North. There is a force of 23 N acting on the ball that has a mass of 1.25 kg. The magnitude of the acceleration due to the force is 23/1.25 = 18.4 m/s^2. The force acts in a direction 30 degrees North of East; the resulting acceleration can be divided into two components one towards the North of 18.4*sin 30 and the other towards the East of 18.4*cos 30 m/s^2.
In 12 seconds the increase in the velocity of the ball is 19 + 18.4*sin 30*12 = 129.4 m/s towards the North and 191.21 m/s towards the East.
The resulting velocity has a magnitude of sqrt(129.4^2 + 191.21^2) = 230.8 m/s^2 in a direction arc tan(129.4/191.21) = 34.08 degrees North of East.
The velocity of the ball after 12 seconds is 230 m/s in a direction 34.08 degrees North of East.