Is the Velocity negative or positive?

AI Thread Summary
The discussion revolves around calculating the velocity of a log thrown over the Grand Canyon, which impacts the ground after 5.65 seconds. The final velocity is calculated as 55.4 m/s, with the sign of the velocity depending on whether gravity is considered positive or negative. Consistency in choosing the sign for acceleration due to gravity (9.8 m/s²) is emphasized, as it affects the final velocity's designation. It is clarified that while gravity is constant, the net force acting on an object can vary, leading to different accelerations in various scenarios. The conversation highlights the importance of unit inclusion and understanding the principles of motion under gravity.
rasofia77
Messages
22
Reaction score
0
Member warned about re-posting the same problem statement in new threads. Infraction points awarded.

Homework Statement


If you were to throw a large log over the edge of the Grand Canyon and it took 5.65 seconds to hit the ground, calculate the velocity of the log at impact in m/s and calculate the distance the log fell in feet.[/B]

Homework Equations


vf=vi+at[/B]

The Attempt at a Solution


So I get vf=(9.8)(5.65)= 55.4 ...But then I think about it, and the object is going downwards so shouldn't it's velocity be negative (South)..? Also, is 9.8 supposed to be positive or negative?[/B]
 
Physics news on Phys.org
You can choose g to be positive or negative, whichever you wish. The final velocity here will take on whatever sign you give to g. The important thing is that you remain consistent! :)
 
NascentOxygen said:
You can choose g to be positive or negative, whichever you wish. The final velocity here will take on whatever sign you give to g.

So there's always two right answers?
 
rasofia77 said:
So there's always two right answers?
Not really. It's final velocity is 55.4 m/s towards the centre of the Earth. That's one answer. :)

I'd be concerned about your omission of the units when talking of velocity, though.
 
NascentOxygen said:
Not really. It's final velocity is 55.4 m/s towards the centre of the Earth. That's one answer. :)

I'd be concerned about your omission of the units when talking of velocity, though.

Sorry, haha *m/s. One last question, is acceleration the same for all objects?
Well, it is, due to gravity (9.8m/s^2) ..but then again I get confused because there's so many different word problems with different accelerations as answers, obviously. What would be the final, true answer though?
 
rasofia77 said:
Sorry, haha *m/s. One last question, is acceleration the same for all objects?
g is the same for all objects in a particular location. Whether the object is accelerating depends on the net force. Gravity is only one of many forces that may be acting simultaneously on an object.

Well, it is, due to gravity (9.8m/s^2) ..but then again I get confused because there's so many different word problems with different accelerations as answers, obviously. What would be the final, true answer though?
The answer to what?
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top