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Urgent: Projectile fired from Earth's surface

  • #1
A projectile is fired vertically from Earth's surface with an initial speed of 8.2 km/s. Neglecting air drag, how far above the surface of Earth will it go?


I am completely stuck. What formula should i use, this was in the Gravitational Potential Energy /escape speed section
 

Answers and Replies

  • #2
609
0
use conservation of energy here...
warning: the potential energy is not mgh in this case
 
  • #3
vince, how do i use conservation of energy if i am not given the mass of the projectile? thanks
 
  • #4
609
0
you don't need the mass, since the KE and the PE is directly proportional to the mass, the mass will cancel out anyway... just put the mass = m in the formulas... the m will not appear in your final answer
 
  • #5
but then if i just put m, i have too many unknowns, i don't know the final velocity, only the initial. so i am stuck
 
  • #6
and what variable should i be solving for if i want to know the final distance from earth?
 
  • #7
Write down the equations first, Cinder, then complain after they don't work.

And if you want to know the final distance from Earth, you should probably be solving for the final distance from Earth.

--J
 
  • #8
I have equations, this is what im using, no need to be so sarcastic, geeze. Sorry if i dont understand it so well.

K(final)+ U(final) = K(initial)+ U(initial)


1/2mvfinal^2 - GMm/R = 1/2 Mvinitial^2 - GMm/R

what i meant was, which one of these variables give the distance from earth? whiich should i solve for?
 
  • #9
609
0
yeah, you have some unknown...
mass of the projectile, which I have already told you it doesn't matter
initial velocity, isn't that given?
final velocity, at the highest point, do you still have velocity?
R initial... your projectile fire from the surface of earth... so R is...........
R final... this is your only unknown... and also your answer... solve for it

and the M... mass of earth... is that an unknown?
 
Last edited:
  • #10
ah crap, i was thinking i didnt have v final, but i guess i had to assume it stops so it is zero. thanks vince, brb leme do my calculations.
 
  • #11
damn how come when i try to isolate R final, the units end up w/ m^2/s^2 * meters, so i end up w/ m/s^2 when i should just get meters for the final radius, correct?


when i isolate Rfinal i get -GMm/(1/2mvinitial^2-Gmm/Riniitial) theni cancel out the masses like u said vince, iand i get

R final = -GM/ (1/2 Vinitial^2- GM/R initial) and the units arent meters, its m/s^2
 
  • #12
609
0
apparently, the unit is fine.....you have GM/(***+GM/R) the GM cancels out and leave the R alone.. how could you get m/s^2??????
and your equation is fine too....
 
  • #13
ok, leme just carry out the number calculations, and see what i get. brb
 
  • #14
in the denominator, I have the kinetic energy still , so my units arrent matchin up to meters on the left hand side.



R final = -GM
------------
(1/2 Vinitial^2 - GM/ Rinitial)
 
  • #15
I see the GM's cancel out in the numerator and denominator. But then i have the 1/2 V initial^2 - 1/ R initial. So that is m/s^2 - meters, i dont get it.
 
  • #16
609
0
what is the unit of GM/v^2... check your data book what is the unit of G
 
  • #17
crap nevermind my algebra was bad! forgot to cancel out the units w/ the first term (kinetic units) in the denomitaor as well, ok brb sorry.
 
  • #18
Vince, since it asks how far about the surface of the earth will it go, do i subtract the radius of the earth from my R final? Cause isnt R final the distance from the middle of the earth to the middle of the projectile.
 
  • #19
Vince, thank you man i got it! I really appreciate your help.
 

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