suppose the water at the top of niagara falls has a horizontal speed of 2.7m/s just before it cascades over the edge of the falls. at what vertical distance below the edge does water reach a horizontal distance of 4 m away
To find the vertical distance below the edge where the water reaches a horizontal distance of 4 m away, we can use the principles of projectile motion.
Step 1: Analyze the given information.
- Initial horizontal speed of the water, Vx = 2.7 m/s
- Horizontal distance the water reaches, Dx = 4 m
Step 2: Find the time taken for the water to reach the horizontal distance.
We can use the formula: Dx = Vx * t
Substituting the given values, we have:
4 m = 2.7 m/s * t
Solving for t, we find:
t = 4 m / (2.7 m/s) ≈ 1.48 s
Step 3: Find the vertical distance traveled during this time.
Using the formula for vertical displacement in free fall:
Dy = (1/2) * g * t^2
Substituting the value of time obtained in Step 1 and the acceleration due to gravity, g = 9.8 m/s^2, we have:
Dy = (1/2) * 9.8 m/s^2 * (1.48 s)^2 ≈ 10.8 m
Therefore, the water reaches a vertical distance of approximately 10.8 m below the edge when it reaches a horizontal distance of 4 m away.