A pipe discharges storm water into a creek. Water flows horizontally out of the pipe at 1.5 m/s, and the end of the pipe is 2.5 m above the creek. How far out from the end of the pipe is the point where the stream of water meets the creek?

### Answer

**Step 1**

Calculate the time at which storm water discharge into a creek.

The expression of the vertical distance in terms of the acceleration due to the gravity is equal to,

`s=u t+\frac{1}{2} g t^{2}`

Substitute for `u`

, `2.5{\rm{ m}}`

for `s`

, and `9.8{\rm{ m/}}{{\rm{s}}^2}`

for `g`

in the above expression of the vertical distance,

`2.5 \mathrm{~m}=0+\left(\frac{1}{2}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\left(t^{2}\right)`

Rearrange the above expression in terms of the time tt .

`t^{2}=\frac{(2)(2.5 \mathrm{~m})}{\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}`

Taken square root both sides,

`t=\sqrt{\frac{(2)(2.5 \mathrm{~m})}{\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}}`

`=0.714 \mathrm{~s}`

**Step 2**

Calculate the point where the stream of water meets the creek.

The expression of the point at which the stream of water meets the creek.

`s=vt`

Substitute `1.5{\rm{ m/s}}`

for `v`

and `0.714{\rm{ s}}`

for tt in the above expression of the `s`

```
\begin{gathered}
s=(1.5 \mathrm{~m} / \mathrm{s})(0.714 \mathrm{~s}) \\
=1.07 \mathrm{~m}
\end{gathered}
```

The point where the stream of water meets the creek is equal to `1.1{\rm{ m}}`

.