Quick Physics 20 Question

[AC]

Should be easy for you guys here, but I am completley stuck on this one workquestion. They supply the answer(3.33 ,/s^2) , but not the solution.

I've tried searching online, and asking about everybody I know.

Anways, if anybody here knows how to solve this, I would forever be in your debt.

When an object is dropped from a height of 20.0m above the surface of Planet Z, it will fall 5.00m during the 2nd second of fall.

What is the acceleration of a falling object near the surface of this planet?

The formulas we are given are:

a = (Vf - Vi)/t

d = ([Vf + Vi]/2)t

d = (Vi)t + .5at^2

Vf^2 = Vi^2 + 2ad

[Mike F]

Originally posted by fotze@May 16 2005, 07:48 PM
5=(0)t+0.5at^2

t=2# solve for a

a= 2.5

there is not enough info if the acceleration is dependant on distance form the earth

That can't be right, because you don't account for the distance travelled during the first second.

OK here you go:

a = (Vf - Vi)/t

Vi=0

Vf -- you know it moves 5m during the second second, so 5m/s is it's average velocity during the second second, which means it's actual velocity is 5m/s half way through the second second. Therefore:

a = (5m/s - 0)/1.5s
a = 5m/s/1.5s
a = 3.33m/s/s

[AC]

Originally posted by Mike F+May 16 2005, 08:22 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Mike F @ May 16 2005, 08:22 PM)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-fotze@May 16 2005, 07:48 PM
5=(0)t+0.5at^2

t=2 solve for a

a= 2.5

there is not enough info if the acceleration is dependant on distance form the earth

That can't be right, because you don't account for the distance travelled during the first second.

OK here you go:

a = (Vf - Vi)/t

Vi=0

Vf -- you know it moves 5m during the second second, so 5m/s is it's average velocity during the second second, which means it's actual velocity is 5m/s half way through the second second. Therefore:

a = (5m/s - 0)/1.5s
a = 5m/s/1.5s
a = 3.33m/s/s [/b][/quote]
Yes! Thankyou!

This damn question has been bugging me all day.

[evman150]

Anthony, I frequent a great physics forum which has a homework help section. If you have further questions the people there are very helpful.

They do not, however, do your homework for you, so you at least have to show you have tried a question before they will help you out. Nevertheless it is a great resource to have when taking a course in Physics/Math/Chemistry/Astronomy.

http://physicsforums.com/index.php?

[shoestring]

Wow I always solve these kind of problems first,but congrats you beet me too it.
:stupid:

Like wow!

[the_only_turek_fan]

Originally posted by Mike F+May 16 2005, 09:22 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Mike F @ May 16 2005, 09:22 PM)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-fotze@May 16 2005, 07:48 PM
5=(0)t+0.5at^2

t=2 solve for a

a= 2.5

there is not enough info if the acceleration is dependant on distance form the earth

That can't be right, because you don't account for the distance travelled during the first second.

OK here you go:

a = (Vf - Vi)/t

Vi=0

Vf -- you know it moves 5m during the second second, so 5m/s is it's average velocity during the second second, which means it's actual velocity is 5m/s half way through the second second. Therefore:

a = (5m/s - 0)/1.5s
a = 5m/s/1.5s
a = 3.33m/s/s [/b][/quote]
fata!!!!

Dude, good work. I was looking at it last night and must have spent about 30 minutes on it and didnt get it.

Like fotze, I have been out of school for too long and am a idiot.

And that has only been 13 months...

[BlackEleven]

Personally, I would have solved it as follows.

First, calculated the inital velocity during the2nd second of falling:
d = (Vi)t + .5at^2
5 = (Vi)*1 + 0.5a(1)(1)
Vi = 5 - 0.5a

Since the inital velocity in the 2nd second of falling is the same as the final velocity in the 1st second of falling, you can subsitute Vi above for Vf in the equation below:
a= (Vf - Vi) / t
a = (5 - 0.5a - 0) / 1
a = 5 - 0.5a
1.5a = 5
a = 3.333

Just offering an alternate solution....

Edited for clarity