Aparticle 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

A) acceleration. this seems to be the one that makes sense.

This is the answer: body 1 + engine 2