The following question has been giving me trouble for a while now, it states
A particle is thrown with speed $10 \space \text{ms}^{-1}$ from a point $2\space \text{m}$ from the ground, at an angle of $45°$ above the horizontal. What is the speed of the particle when it is at a height of $4 \space \text{m}$ for the second time in its motion?
I have tried the following,
$$v^2-u^2=2as \iff (v\sin(45))^2=(10\sin(45))^2+2(-10)(2)$$ $$v=\frac{\sqrt{10}}{\sin(45)} \implies v=4.47\space \text{ms}^{-1} $$
But I did not get the correct answer of $v=7.75\space \text{ms}^{-1}$, similarly I tried using the equation $s=\frac{1}{2}at^2+ut$ to first find the time taken and then plug it into $v=at + u$ to yield the velocity however this was to no avail.
Where am I going about wrong, I have a sneaking suspicion that it is to do with the value of $s$.
In the horizontal direction, where you have no acceleration, $v_x=v_{x0}=v_0\cos45^\circ$. In the vertical direction $$v_y^2=v_{y0}^2-2gh=v_0^2\sin45^\circ-2gh$$ Then $$v^2=v_x^2+v_y^2=v_0^2-2gh$$ Notice that the angle does not matter. You could have gotten this from the conservation of energy.