Trajectory with Force Field

Since the particle isn't initially in the force field, its equation of motion is

$\vec{x} = (-10, 0)^T m + (10, 10)^T \frac{m}{s} \cdot t + \frac{1}{2} (0, -10)^T \frac{m}{s^2} \cdot t^2$ .

The y-position doesn't matter, but it's obvious that it will take 1 second to reach $x=0$ . The y-velocity at this point is

$v_y( t= 1s) = 10 \frac{m}{s} - 10 \frac{m}{s^2} \cdot 1s = 0$ ,

while the x-velocity didn't change. Now, to calculate the time it takes to reach $x=0$ again:

$0 = 0 + 10 \frac{m}{s} t - \frac{1}{2} \cdot 5 \frac{m}{s^2}t^2$

$t = 4s$

$t = 0$ would also solve the above equation, but since it is the point of entry into the force field, 4s is the point of exit we are looking for. The force field of 5N on a 1kg weight will cause $5\frac{m}{s^2}$ of acceleration in $-x$ - direction, so the velocity at exit is:

$\vec{v}(4s) = (10, 0)^T\frac{m}{s} + (-5, -10)^T \frac{m}{s} \cdot 4s$

$= (-10, -40)^T \frac{m}{s}$

The magnitude of this vector is $\sqrt{ 10^2 + 40^2 } = 41.231$ .