Formula for free-falling object?

[J-Girl]

Hi there, my question has to do with the velocity of a free falling object. I know that V=Vo + at, but if you don't 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. I am 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!

 

[Hootenanny]

Hi there, my question has to do with the velocity of a free falling object. I know that V=Vo + at, but if you don't 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. I am 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.

The final equation you used is formally the average velocity - but it is fine to use in problems where the acceleration is constant.

 

[J-Girl]

oh cool! Thanks:) Also, the rest of the question was based on the kinetic energy and the momentum. I haven't 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, I am 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:)

 

[Hootenanny]

oh cool! Thanks:) Also, the rest of the question was based on the kinetic energy and the momentum. I haven't 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, I am 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.

 

[J-Girl]

Thanks:) :)

 

[Hootenanny]

Thanks:) :)

A pleasure