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)at²10 = 0 + (1.6)t + (1/2)(0)t²10/1.6 = tt = 6.25sSo 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 = ma8 = (2.5)aa = 3.2m/s²Now, use kinematics again.Y = y + vt + (1/2)at²Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²Y = 62.5m