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}}.
