On a day when the water is flowing relatively gently, water in the niagara river is moving horizontally at 4.5 m/s before shooting over niagara falls. after moving over the edge, the water drops 53 m to the water below.

I think below are the questions:

1st question

If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom?

How far does the water move horizontally during this time?

Below is the answers:

D = Dinitial + Vinitial*t + (1/2) * a*t^2

Let's assume up is positive, and the bottom of the falls is zero displacement. So you get

0 = 53 + 0*t + (1/2) * (-9.8) * t^2 = 53 - 4.9*t^2

you can then solve for t

t = sqrt (53/9.8) or about 2.3 s

second question

D = Dinitial + Vinitial*t + (1/2) * a*t^2

D = 0 + 4.5 * sqrt (53/9.8) + 0*t^2 = 4.5*sqrt (53/9.8) or about 10.5 m