Suppose the water at the top of Niagara Falls has a horizontal speed of 2.36 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 38.8 ° angle below the horizontal?

Question

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.36 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 38.8 ° angle below the horizontal?

Answer 1

To find the vertical distance below the edge where the velocity vector of the water points downward at a specific angle, we can use trigonometry and the principles of projectile motion.

First, let's break down the given information:
- The horizontal speed of the water just before it cascades over the edge is 2.36 m/s.
- The angle at which the velocity vector points downward is 38.8° below the horizontal.

We can start by determining the horizontal and vertical components of the velocity.

Horizontal Component (Vx):
The horizontal speed gives us the magnitude of the horizontal component.
Vx = 2.36 m/s

Vertical Component (Vy):
To find the vertical component, we will use trigonometry.
We can consider the angle of 38.8° below the horizontal as the angle of elevation.
The vertical component can then be found using the sine of the angle:
Vy = V * sin(θ)
Vy = 2.36 m/s * sin(38.8°)

Next, we can determine the time it takes for the water to fall from the top to the point where the velocity vector points downward.

The vertical distance traveled by the water can be calculated using the formula:
d = (1/2) * g * t^2
where 'd' is the vertical distance, 'g' is the acceleration due to gravity (approximately 9.8 m/s²), and 't' is the time taken.

Rearranging the formula gives us:
t = sqrt(2d / g)

Since the water starts with an initial vertical velocity of 0 m/s, the time taken to fall from the top to the given point is the same as the time taken for the vertical component of the velocity to reach its maximum value.

Now we can equate the vertical component of the velocity to the maximum value reached during free fall.

Vy = g * t
2.36 m/s * sin(38.8°) = 9.8 m/s² * t

Using this equation, we can solve for t.

Finally, we can substitute the calculated value of 't' back into the equation for 'd' to find the vertical distance below the edge where the velocity vector points downward.

d = (1/2) * g * t^2

This will give us the answer to the original question.

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.36 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 38.8 ° angle below the horizontal?

95

To find the vertical distance below the edge where the velocity vector of the water points downward at a specific angle, we can use trigonometry and the principles of projectile motion.