Question

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.73 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 52.9 ° angle below the horizontal?

Answer 1

required vertical distance below the edge is 0.6648 m

Step-by-step explanation:

Given the data in the question;

Horizontal speed of water falls v = 2.73 m/s

direction of water falls 52.9° below the horizontal

The vertical velocity must be such that;

tanθ = v
`_y` / v
`_x`

Now, vertical speed of water falls;

v
`_y` = v
`_x` × tanθ

we substitute

v
`_y` = 2.73 × tan(52.9°)

v
`_y` = 2.73 × 1.322237

v
`_y` = 3.6097

Now, at the top of falls, initial speed u = 0

v² - u² = 2as

s = ( v² - u² ) / 2as

we substitute

s = ( 0² - (3.6097)² ) / (2 × 9.8)

s = 13.029934 / 19.6

s = 0.6648 m

Therefore, required vertical distance below the edge is 0.6648 m