91Ó°ÊÓ

The distance travelled by the front of the flow, d = 920m

The angle made by the slope with horizontal, $\mathrm{θ}=10°$

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

Find theinitial speed of the front of the flowusing the law of conservation energy.

Formula:

$U_i+K{E}_i=U_f+K{E}_f$

Here, $U_i$is the initial potential energy, $K{E}_i$is the initial kinetic energy, $U_f$is the final potential energy, and $K{E}_f$is the final kinetic energy.

According to the conservation of energy,

$$\begin{gathered} U_i+K{E}_i=U_f+K{E}_f \\ mgh=\frac{1}{2}m{v}^2 \end{gathered}$$

Since the slope makes $10°$angle with the horizontal,

$$\begin{gathered} mgd\sin \mathrm{θ}=\frac{1}{2}m{v}^2 \\ {v}^2=2gd\sin \mathrm{θ} \\ v=\sqrt{2gd\sin \mathrm{θ}} \\ v=\sqrt{2(9.8\mathrm{m}/{\mathrm{s}}^2)(920\mathrm{m})\mathrm{s}9\mathrm{n}10°} \\ =\sqrt{(18,032{\mathrm{m}}^2/{\mathrm{s}}^2)(0.174)} \\ =\sqrt{3,137.568{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =56\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the initial speed of the front of the flow is 56 m/s