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emargarida Rookie
Joined: 17 Dec 2010 Posts: 7 Location: London

Posted: Mon Jul 25, 2011 3:17 am Post subject: Can anyone help me with this physics problem? 


Hi,
I seen unable to solve this problem, could anyone help me?
"An object is dropped from a height h and strikes ground with a velocity v. If the object is dropped from a height of 2h, which of the following represents its velocity when it strikes the ground?"
Thanks 

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emargarida Rookie
Joined: 17 Dec 2010 Posts: 7 Location: London

Posted: Mon Jul 25, 2011 3:19 am Post subject: 


the options go:
a) v
b) 1.4v
c) 2v
d) 4v
Thanks 

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cat2devnull Regular
Joined: 23 Jul 2010 Posts: 23 Location: Adelaide

Posted: Mon Jul 25, 2011 11:49 am Post subject: 


You need to have your head around the equations of motion so start by looking them up on wikipedia and also look at the Khan Academy projectile motion videos.
v^2 = u^2 + 2as
where
s = the distance between initial and final positions (height)
u = the initial velocity = 0
v = the final velocity
a = the constant acceleration (gravity)
Thus:
v^2=2as
or another way of looking at it is:
v^2=2gh (gravity and height)
or
v=√2gh
So if you double the height then you get;
v=√4gh
Aka you multiply v by √2 or 1.4 hence option b.
You can actually calculate this properly by building the two equations and substituting one into the other using the common var of gravity but you will run out of time in the exam if you actually were to do this.
v^2=2gh
g=v^2/2h
thus if you look at the two scenarios (using capital v and h for the second experiment where H=2h);
g=v^2/2h and g=V^2/2H
so
v^2/2h = V^2/2H
since H=2h
v^2/2h = V^2/4h
now you can cancel out the h
v^2/2 = V^2/4
multiply both sides by 4 to get rid of the denominator
2v^2 = V^2
square root both sides to get rid of the square
V = v√2
Ta da! the new Velocity is √2 (or 1.4) times the old velocity. 

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emargarida Rookie
Joined: 17 Dec 2010 Posts: 7 Location: London

Posted: Mon Jul 25, 2011 8:40 pm Post subject: 


thank you very much!!! That was extremely helpful. 

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