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?

required vertical distance below the edge is 0.6648 m

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 / v

Now, vertical speed of water falls;

v = v × tanθ

we substitute

v = 2.73 × tan(52.9°)

v = 2.73 × 1.322237

v = 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

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

a, c, d, e are all applying about success strategies.