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.

Respuesta :

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

Answer:

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 answer:  

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

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