• Support PF! Buy your school textbooks, materials and every day products Here!

Formula for free-falling object?

  • Thread starter J-Girl
  • Start date
  • #1
38
0
Hi there, my question has to do with the velocity of a free falling object. I know that V=Vo + at, but if you dont know the time and the final velocity of the object, can you use another equation? I found this one"V^2=Vo^2 + 2(as)"-s being the distance travelled. Im trying to find the final velocity of an object that fell 134 meters from a bridge. Can somebody tell me please, if this is right, or point me in the right direction?

Distance=134m
Acceleration=9.8m.s^-2

So I did:
V^2=Vo^2 + 2(as)
= V^2=0^2 + 2(9.8 x 134)
=V^2=2(1313.2
=V^2=2626.4
v=51.25 m per second
also i need to know the time, so i did:
t=(v-vo)/a
t=(51.25)/9.8
t= 5.23 seconds
im just not quite sure..also, is this equation for the average velocity or for the final velocity? thanks!!
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Hi there, my question has to do with the velocity of a free falling object. I know that V=Vo + at, but if you dont know the time and the final velocity of the object, can you use another equation? I found this one"V^2=Vo^2 + 2(as)"-s being the distance travelled. Im trying to find the final velocity of an object that fell 134 meters from a bridge. Can somebody tell me please, if this is right, or point me in the right direction?

Distance=134m
Acceleration=9.8m.s^-2

So I did:
V^2=Vo^2 + 2(as)
= V^2=0^2 + 2(9.8 x 134)
=V^2=2(1313.2
=V^2=2626.4
v=51.25 m per second
also i need to know the time, so i did:
t=(v-vo)/a
t=(51.25)/9.8
t= 5.23 seconds
im just not quite sure..also, is this equation for the average velocity or for the final velocity? thanks!!
Spot on. :approve:

The final equation you used is formally the average velocity - but it is fine to use in problems where the acceleration is constant.
 
  • #3
38
0
oh cool! Thanks:) Also, the rest of the question was based on the kinetic energy and the momentum. I havent really done a lot of that so far, I have been trying to figure out velocity and acceleration equations! I think I have the right formulas, im just not sure how to convert my final answer.
3. "Calculate the momentum of the person just before they hit the water" so if p=mv, then is the equation simply :
p=mass(which is 70kg)x final velocity(51.25)
p=70kg x 51.25m/s^-1
and the to get the answer in kilograms per meter per second, i divided by 51.25 to get
p=1.366kg.m.s^-1
is this the right format to write an answer where there are two units of measurement?

the last part of the question was:
4. "Calculate the kinetic energy of the person, just before they hit the water"
i found a formula which was KE=1/2(mv^2)
so i did KE= 1/2(70kg x 51.25^2)
KE=1/2(70kg x 2626.53m.s^-1)
KE=35kg x 1313.265m.s^1
and then i divided by 1313.265, so that i could get it in kilograms per meter per second, and my final answer was:
KE= 0.02665kg.m.s^-1
KE= 2.665 x 10^-2 kg.m.s^-1
soo, not sure if that was all jibberish or not lol, please get back to meee:)
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
oh cool! Thanks:) Also, the rest of the question was based on the kinetic energy and the momentum. I havent really done a lot of that so far, I have been trying to figure out velocity and acceleration equations! I think I have the right formulas, im just not sure how to convert my final answer.
3. "Calculate the momentum of the person just before they hit the water" so if p=mv, then is the equation simply :
p=mass(which is 70kg)x final velocity(51.25)
p=70kg x 51.25m/s^-1
This is correct.
the last part of the question was:
4. "Calculate the kinetic energy of the person, just before they hit the water"
i found a formula which was KE=1/2(mv^2)
so i did KE= 1/2(70kg x 51.25^2)
KE=1/2(70kg x 2626.53m.s^-1)
KE=35kg x 1313.265m.s^1
This is also correct.

There is no need to do additional conversions or calculations. You're quantities are already in appropriate (S.I.) units: kg.m/s for momentum and kg.m2/s2 = Joules for kinetic energy.
 
  • #5
38
0
Thanks:) :)
 
  • #6
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6

Related Threads for: Formula for free-falling object?

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
352
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
848
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
12
Views
14K
Top