A particle that has a mass of 2.5 kg is moving in the positive X-direction with a constant velocity of 1.6m/s. Suddenly a constant vertical force with a magnitude of 8N is applied in the positive Y-direction. what is the resulting vertical displacement over the time taken for a 10m horizontal displacement?

The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.

X = x + vt + (1/2) at²

10 = 0 + (1.6) t + (1/2) (0) t²

10/1.6 = t

t = 6.25s

So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.

Start with Newton's second law.

F = ma

8 = (2.5) a

a = 3.2m/s²

Now, use kinematics again.

Y = y + vt + (1/2) at²

Y = 0 + (0) (6.25) + (1/2) (3.2) (6.25) ²

Y = 62.5m