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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 $\theta$.

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I think you are overcomplicating things. You can consider the horizontal motion (no acceleration, no air resistance) separately from the vertical motion. The initial vertical speed is 0 m/s, and you know the acceleration and the distance travelled. Use this to find the time taken for the water to drop. The same time will apply to the horizontal motion, and you know the distance, so can compute the velocity. –  Mark Bennet Jun 20 '12 at 13:35

2 Answers 2

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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 $v$.

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Split it into horizontal and vertical components, with

$$ s = ut + {at^2 \over 2} $$

you can calculate the time it takes to hit the sea in vertical plane, then plug this back into

$$ speed = {dist \over time} $$

to find the initial speed as the horizontal speed component never changes.

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