opted

[eyetime]

a ball is thrown horizontally at 20 m/s from the top of a building and strikes the level ground 50 m from the building. Approximately how many meters high is the roof of the building?

How can you figure that out with this little information, wouldnt I need time to plug into the equation and calculate 'h'?

and a 50 kg girl pulls a 20 kg wagon with a force of 10 N. The wagon accelerates 2m/s. what is the force exerted by the wagon on the girl?

I chose 10 which is correct because of newtons 3rd law, but is there an equation to prove it?

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[Arctic]

Since it is a free, publicly available practice test, I have no reservations about discussing those problems.

And please check out my free OAT study articles on www.optometrystudents.com

For the first problem, you know that there is NO horizontal acceleration. There is only the vertical acceleration caused by gravity. Therefore, the horizontal velocity, when coupled with the horizontal distance traveled, can give you the flight time, and you can temporarily ignore the height of the building. Flight time is just 50/20= 2.5 seconds. Once you know this, it is very easy to calculate the height, h, of the building, because you know that, starting from rest, the ball dropped took 2.5 seconds to hit the ground (this is completely independent of the ball's horizontal velocity).
The simplest equation to use here is d= vi t + 1/2 a t^2

You know vi, the initial velocity (remember, this is for the vertical component of velocity) is zero, so you have d= 1/2 a t^2 (if anyone cares, I will gladly explain, in words, how you derive this equation for distance)
This becomes d= 1/2 x 10 x 2.5^2 and because you don't have a calculator, 2.5^2= 25/4 = 6.25 and 6.25 x 10 is obviously 62.5, and half of that is just 31.25
Obviously this isn't an absolute exact answer, as I used 10m/s^2 for gravitational acceleration instead of 9.81m/s^2 or whatever it is.


For the second problem, I'm not exactly sure what you are asking. You know that the girl is pulling the wagon with a force of 10N and, yes, the equal and opposite force to that is going to be 10N from the wagon on the girl (you know whenever you pull on something heavy enough, that you can feel the straining force in your arm, and it feels a bit like that object is pulling back on your arm). Asking for an equation to prove a physical law doesn't exactly make sense to me, because the equations are the mathematical manifestations of the physical laws. Equations aren't the end all to the universe, they are just the 'end all' to math, which is a tool that we, as humans, created to help us describe our world. That's just my opinion, anyways.

 

[EyEnStein 07]

I chose 10 which is correct because of newtons 3rd law, but is there an equation to prove it?

Ofcourse there is an equation to prove it...its exactly as you stated, Newtons Third Law:

F12 = -F21

where 12 denotes force of one due to two, or force on two exerted by one. Its actually whatever convention you take it to be. Remember that F is a vector quantity...so the negative sign doesn't mean the value is negative, it tells you the direction.


Lastly, check this out it might be of help to you for other questions you may have:
http://forums.studentdoctor.net/showthread.php?t=574957

 

[Arctic]

That's like saying the equation for gravitational attraction proves gravity. I don't really care about a 'philosophical' argument or anything, but an equation that IS the equation OF a law certainly doesn't prove the law. Rather, the experimental evidence is what 'proves' the law.

 

[eyetime]

I only asked for the equations because they gave so much extra information. I wasn't sure if I arrived at the right answer because I'm a decent guesser, or if there was a mathematical component behind it. Thank you for all your help!!