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:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.