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Homework Help: Imagine that a bullet is shot vertically into the air with an initial

  1. Jun 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Imagine that a bullet is shot vertically into the air with an initial speed of 9800 m/s. If we ignore air friction, how high will it go?

    vi = 9800 m/s
    h = ?

    2. Relevant equations

    v2 = vi2 + 2aΔy
    Ki = Ui --> 1/2mv2 = mgh

    F = (Gm1m2)/r

    3. The attempt at a solution

    Attempt 1

    v2 = vi2 + 2aΔy
    0 = (9800 m/s)2 + 2(-9.8 m/s2)(Δy)
    Δy = 4,900,000 m

    Attempt 2

    1/2mv2 = mgh
    h = 4,900,000 m

    Neither of these equations worked, so I tried using gravitational equations, but I couldn't figure it out.
  2. jcsd
  3. Jun 18, 2014 #2


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    Do you know calculus?
    What's the actual answer given in the text book?
  4. Jun 18, 2014 #3

    Unfortunately, I do not know calculus, and there is no answer listed in the back of the book.
  5. Jun 18, 2014 #4


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    Your answer is correct if we assume that acceleration is constant.Who said it's wrong?

    If you don't know calculus, then you should not bother about the varying acceleration..
    Edit:It seems you should. What's your grade?
    Last edited: Jun 18, 2014
  6. Jun 18, 2014 #5
    I typed it into an online program called CAPA, and it replied that my answer was incorrect. I just emailed my professor, and he said that I need to use the full gravitational potential energy equation because the gravitational force is not constant. I've tried a few more attempts, but none of them have been correct.
  7. Jun 18, 2014 #6


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    Please show your attempts then
  8. Jun 18, 2014 #7


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    Yes, the value you get for Δy should tell you to use the full gravitational force equation since the value you get is of the order of magnitude of the radius of the earth. So the force will be a function of height and you cannot use the approximation U=mgh.
  9. Jun 18, 2014 #8
    Sorry, just got back home.

    First, I tried using this equation: 1/2mv2 = (Gm1m2/r2)Δy
    Δy = (v2 * r2)/2Gm1
    Δy = (98002* 63800002) / (2*(6.67*10-11)* (5.97*1024))
    Δy= 4908664 m

    Next, I tried omitting the Δy, instead using: r = sqrt(Gm/v2)
    r = sqrt((6.67*10-11)* (5.97*1024)/ 98002)
    r = 2036 m
  10. Jun 18, 2014 #9
    How can I set up my equation so that force is a function of height? I've attempted to work in G and Earth's mass and radius, but I attained an answer too close to my previously found incorrect answers.
  11. Jun 18, 2014 #10


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    Your first attempt above is incorrect because you are again using the fact that F=mg, which is only true when the height of projectile << radius of earth. Formally, to show that we do a Taylor expansion.

    I am not quite sure what you did in the second attempt by simply throwing away some terms.

    The method I had in mind does involve calculus as adjacent alluded to. The starting point is to write $$m_1 \ddot{z} = -\frac{Gm_1 m_2}{r^2} = -\frac{Gm_1 m_2}{(r_e + z)^2}$$ and then to do some chain rule and integration.
  12. Jun 18, 2014 #11
    Hmm... How would I incorporate the velocity of the object then?

    I tried using 1/2mv2 = -(Gm1m2/r), but ended up getting a very small number.
    98002 = - (2*(6.67*10-11)*(5.97*1024))/(6.4*106 - h)2

    I ended up getting 6402880 m, but I don't want to enter it into the program because I have used almost all of my attempts already.
  13. Jun 18, 2014 #12


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    The basic idea you want to use is that kinetic energy + potential energy is conserved. Use the correct potential function and it should all fall into place. I'm not a physicist but this is surely how it works.

    I think this should not be complicated.
  14. Jun 18, 2014 #13
    I know it shouldn't be complicated, and that's making it all the more frustrating. I can't seem to attain the correct answer after spending hours on this one question. I only have two attempts left.

    I tried doing 1/2mv2 + mgh = - (GmM/r), so that
    1/2(9800)2 + (9.8 m/s2)h = -((6.67*10-11)(5.97*1024)/6380000
    h = 1.13 * 107 m (but that is incorrect as well)

    I would really appreciate if someone could hold my hand and take me through this because I have no clue what to do.
  15. Jun 18, 2014 #14

    Doc Al

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    You're mixing up two equations for the the potential energy. Stick to the full one. Hint: r will vary from Rearth to Rearth+h.
  16. Jun 18, 2014 #15


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    Try using the full expression for the potential both for the initial and for the turning point. At what r are you at the initial point? At what r are you at the turning point?
  17. Jun 18, 2014 #16
    mgh is the potential energy above the surface if the force of gravity is assumed constant, so you can't use that.

    -(gmM/r) is the potential energy of the bullet compared to the potential energy at infinity.
    You need the potential energy with respect to the ground.
  18. Jun 18, 2014 #17
    Oooh... So the equation needs to be set up as 1/2mv2 + (-GmM/r) = (-GmM/r+h)

    Thank you to everyone who helped!!! I really appreciate your cooperation with such a dunce as myself! :p
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