Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

Question

05-01-2024
Physics

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Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

Answer :

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jacob193 jacob193 · Answer 1 · 2024-01-05T21:39:09-05:00

Answer:

Approximately [tex]5.2\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance on the water particles is negligible.)

Explanation:

Under the assumptions, the water particles would accelerate at a constant [tex](-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] (downward) in the vertical component. Velocity of the water particles in the horizontal component would be constantly [tex]2.7\; {\rm m\cdot s^{-1}}[/tex].

In this question, the goal is to find the distance travelled in the vertical direction when the velocity vector points at [tex]75^{\circ}[/tex] below the horizon.

Observe that since the horizontal component ([tex]x[/tex]-component) of velocity is known, the vertical component ([tex]y[/tex]-component) of velocity at the given moment can be found using trigonometry:

[tex]\begin{aligned} & (\text{velocity, $y$}) \\ =\; & (\text{velocity, $x$})\, \frac{(\text{``opposite" to angle of elevation})}{(\text{``adjacent" to angle of elevation})} \\ =\; & (\text{velocity, $x$})\, \tan(\text{angle of elevation}) \\ =\; & (2.7 \; {\rm m\cdot s^{-1}})\, \tan(-75^{\circ})\\ \approx\; & (-10.077)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

(Negative because velocity in the vertical component points downward.)

Given that the water was initially travelling horizontally, the initial velocity in the vertical component would be [tex]0\; {\rm m\cdot s^{-1}}[/tex]. Apply the following SUVAT equation to find the displacement in the vertical component:

[tex]\displaystyle x_{y} = \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a_{y}}[/tex],

Where:

[tex]x_{y}[/tex] is the displacement in the vertical component,
[tex]v_{y} = (-10.077)\; {\rm m\cdot s^{-1}}[/tex] is the current velocity in the vertical component,
[tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] is the initial velocity in the vertical component, and
[tex]a_{y} = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] is the acceleration in the vertical component.

Evaluate to find the value of [tex]x_{y}[/tex]:

[tex]\begin{aligned} x_{y} &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a_{y}} \\ &\approx \frac{(-10.077)^{2} - (0)^{2}}{2\, (-9.81)}\; {\rm m} \\ &\approx (-5.2)\; {\rm m}\end{aligned}[/tex].

In other words, the velocity vector would point at [tex]75^{\circ}[/tex] below the horizon when the water particles are at approximately [tex]5.2\; {\rm m}[/tex] below the initial position (hence the negative sign in front of displacement.) These particles would have travelled approximately [tex]5.2\; {\rm m}\![/tex] in the vertical direction to reach that position.