Suppose the water at the top of Niagara Falls has a horizontal speed of 2.59 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 67.9 ° angle below the horizontal?
Vx = 2.59
Vy = 9.8t
distance fallen = 4.9t^2
tan67.9° = 2.46 = Vy/Vx, so Vy = 2.46*2.59 = 6.37
so, the distance fallen is 4.9(6.37/9.8)^2 = 2.07 m
To solve this problem, we need to find the vertical distance below the edge where the velocity vector is pointing downward at a specific angle below the horizontal.
First, let's identify the given information:
- The horizontal speed of the water just before it cascades over the edge is 2.59 m/s.
- The angle that the velocity vector points downward below the horizontal is 67.9 degrees.
To find the vertical distance, we can use the trigonometric relationship between the horizontal and vertical components of the velocity.
The horizontal component of the velocity can be found using the formula:
horizontal velocity = speed * cosine(angle)
horizontal velocity = 2.59 m/s * cos(67.9°)
horizontal velocity ≈ 1.05 m/s
The vertical component of the velocity can be found using the formula:
vertical velocity = speed * sine(angle)
vertical velocity = 2.59 m/s * sin(67.9°)
vertical velocity ≈ 2.35 m/s
Now, let's consider the motion of the water as it falls over the edge of Niagara Falls. The only force acting on the water at this point is gravity, which causes it to accelerate downward.
Using the equations of motion, we can calculate the vertical distance traveled by the water when its velocity vector points downward at the specified angle.
The equation to calculate the vertical distance covered by an object under constant acceleration is:
vertical distance = initial velocity * time + 0.5 * acceleration * time^2
Since we want to find the vertical distance below the edge, we know the initial velocity is 0 m/s (as the water isn't moving vertically at the edge), and the acceleration due to gravity is approximately 9.8 m/s^2.
By rearranging the equation, we can solve for time:
0 = 0 * t + 0.5 * 9.8 * t^2
4.9 * t^2 = 0
t^2 = 0
From this equation, we can see that the time it takes for the downward velocity of the water to reach the specified angle is 0 seconds.
Since the time taken is 0 seconds, the vertical distance covered by the water is also 0 meters.
Therefore, the vertical distance below the edge where the velocity vector of the water points downward at a 67.9° angle below the horizontal is 0 meters.