so for the life of me i cant see where im going wrong with this question any help appreciated.
The question is that your given two forces that act on a particle of mass $2.4 kg$ one of them is the gravitational force so $$F_g=-mge_y$$ ($e_y$ being the unit vector in the y direction and with $g=-9.8$)
We are told the particle is initially at rest but after 2.2 seconds it has moved $(3.6e_x,-3.5e_y)$ meters.
And we have to find the unknown force $F_u$
So here's what I've done so far, I know that $F_{total}=F_u+F_g$ (I know the question doesn't say we are in equilibrium but the lecturer hasn't done systems not in equilibrium yet) so we have the velocity at 2.2 seconds which is $(\frac{3.6}{2.2}e_x+\frac{-3.5}{2.2}e_y)$ so the acceleration must be $(\frac{3.6}{(2.2)^2}e_x+\frac{-3.5}{(2.2)^2}e_y)$ since we started at rest and accleartion is just rate of change of velocity. Then using the $F=ma$ and the first equation I found that $$F_u=\frac{216}{121}e_x+21.78446281 e_y$$ But this is wrong and I have no idea why.
You are confusing mean velocity with instant velocity. Dividing $3.6$ and $-3.5$ by $2.2$ gives you the mean velocity in both $x$ and $y$ directions (respectively), but they are not constant along those $2.2$ seconds, since the particle is being accelerated.
Consider the displacement of the particle: We have $\Delta x = a_xt²/2$ and $\Delta y = a_yt²/2$, where $a_x$ and $a_y$ are the resulting acceleration in the horizontal and vertical axis (resp.). So $3.6 = a_x(2.2)²/2$ and $-3.5 = a_y(2.2)²/2$, which gives us $a_x = 3.6/2.42 \approx 1.4876 \, m/s^2$ and $a_y = -3.5/2.42 \approx -1.44628 \, m/s^2 $.
Note that $a_y = a_{u_y} + g = a_{u_y} - 9.8$, which gives us $a_{u_y} = 8.35372 \, m/s^2$. Therefore $\vec{F_u} = ma_{u_x}e_x + ma_{u_y}e_y = (2.4)(1.4876)e_x + (2.4)(8.35372)e_y = 3.57024e_x + 20.048928e_y$ $N$ and magnitude $F_u = \sqrt{3.57024² + 20.048928²} \approx 20.3643 N$.