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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.