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 Following are the issues:  

First question  

In the event that we disregard air opposition, what amount of time does it take for the water to go from the highest point of the tumbles to the base?  

How far does the water move evenly during this time?  

The following is the appropriate  

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

How about we accept up is certain, and the base of the falls is zero dislodging. So you get  

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

you would then be able to understand 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  

Related term:

Air resistance:

Air resistance terms the forces that are in resistance to the comparative motion of an object as it permits through the air. These drag forces act differing to the advancing flow velocity, thus slowing the object down.

Answer details:

Subject: Physics

Level: High school

Keywords:

•Resistance

•Air resistance

•Horizontally

Learn more to evaluate:

i think it would be 5