Suppose the water near the top of Niagara Falls has a horizontal speed of 2.3 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 55° angle below the horizontal?
I'm not sure what do with the angle.
That would be where the vertical velocity component Vy is related to the horizontal component Vx by
Vy/Vx = tan 55 = 1.428
Vy = 3.285 m/s (downward)
Calculate how far it must fall (Y) to reach that vertical velcoity component.
Vy = sqrt(2 g Y) = 3.285 m/s
2 g Y = 10.79 m^2/s^2
Y is less than 1 meter!
To solve this problem, we can use trigonometry to find the vertical distance below the edge where the velocity vector of the water points downward at a 55° angle below the horizontal.
Step 1: Break down the given information:
- The horizontal speed of the water just before it cascades over the edge is 2.3 m/s.
- The angle measured from the horizontal is 55°.
Step 2: Calculate the vertical component of the velocity:
Using trigonometry, we can find the vertical component of the velocity by multiplying the horizontal speed by the sine of the angle:
Vertical component of velocity = Horizontal speed * sine(angle)
Vertical component of velocity = 2.3 m/s * sin(55°)
Step 3: Calculate the vertical distance below the edge:
The vertical distance below the edge can be calculated using the equation for constant acceleration:
Vertical distance = (Vertical component of velocity)^2 / (2 * acceleration due to gravity)
Acceleration due to gravity near Earth's surface is approximately 9.8 m/s^2.
Vertical distance = (Vertical component of velocity)^2 / (2 * 9.8 m/s^2)
Step 4: Calculate the vertical distance below the edge:
Substitute the value we calculated for the vertical component of velocity into the equation:
Vertical distance = (2.3 m/s * sin(55°))^2 / (2 * 9.8 m/s^2)
Vertical distance = (2.3 m/s * 0.819)^2 / (2 * 9.8 m/s^2)
Vertical distance ≈ 0.748 m or 74.8 cm
Therefore, the vertical distance below the edge where the velocity vector of the water points downward at a 55° angle below the horizontal is approximately 0.748 meters or 74.8 centimeters.
To find the vertical distance below the edge at which the velocity vector of the water points downward at a 55° angle below the horizontal, we need to use trigonometry.
First, let's break down the given information:
Horizontal speed of the water (v_horizontal) = 2.3 m/s
Angle below the horizontal (θ) = 55°
We can assume that there is no significant change in speed of the water as it falls over the edge, so the only change is in its direction.
To calculate the vertical velocity (v_vertical), we can use the trigonometric relationship:
v_vertical = v_horizontal * sin(θ)
Substituting the given values:
v_vertical = 2.3 m/s * sin(55°)
Calculating the value:
v_vertical = 2.3 m/s * 0.81915
v_vertical ≈ 1.883 m/s
Now, we need to find the vertical distance (d_vertical) the water travels in order for its velocity vector to be point downward at a 55° angle below the horizontal.
We can use the kinematic equation:
d_vertical = (v_vertical^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values:
d_vertical = (1.883 m/s)^2 / (2 * 9.8 m/s^2)
Calculating the value:
d_vertical ≈ 0.181 m
Therefore, the vertical distance below the edge at which the velocity vector of the water points downward at a 55° angle below the horizontal is approximately 0.181 meters.