I would please like help on the following question related to projectile motion.
A horizontal drainpipe 6 metres above sea level empties stormwater into the sea. If the water comes out horizontally and reaches the sea 2 metres out from the pipe, find the initial velocity of the water, correct to 1 decimal place. Let g be 10 ms−2 and neglect air resistance.
I have figured out that when $t=0$, $y=6$. $x=2$ is the range of the projectile motion so when $x=2$, $y=0$.
I have also found the equations for velocity and displacement $$\dot{x}=v\cos\theta$$ $$x=vt\cos\theta$$ $$\dot{y}=-10t+v\sin \theta$$ $$y=-5t^{2}+vt\sin\theta+6$$
I know that to find the range of the particle you set $y=0$ and make $t$ the subject then sub $t$ into $x$.
So my equation is now: $$0=-5t^{2}+vt\sin\theta+6$$
Here's where I get stuck. I'm not sure what to do as both $\theta$ and $v$ are unknown. I was thinking of using the quadratic equation but then rethought it as I still don't have the value of $\theta$.
The vertical speed component when the water exits the pipe is zero, then for the vertical movement
$$ s_v = \dfrac{a \cdot t^2}{2} $$
You know $a$ and $s_v$, so you can find $t$. Then, since horizontal speed is constant,
$$ s_h = v \cdot t $$
You just calculated $t$, and you know $s_h$, so you can find $v$.