Question

Suppose the water at the top of Niagara Falls has a horizontal speed of 3.52 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 78.3 ° angle below the horizontal?

Answer 1

Sol. :- We are given initial velocity along X-axis V_x= 3.2m/s

and let velocity along Y-axis = V_Y

So we have velocity V = Vx  + V_Y

So V_y should be such that it create angle with edge as shown in diagram

So V_y/V_x

So V_y = V_x

So V_y is around 17m/s .

We know that initial velocity along y axis is zero as water was moving only in forward direction just before falling from edge.

So using newton's equation of motion

V² - U² = 2aS , for vertical axis a = gravitational pull = 9.8m/s² and U = 0

So S = V²/2g = = 14.74m

So at around 14.74 m of vertical height water will be at below the horizontal.

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