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A bullet falling to earth and g being the only pull

  1. Dec 9, 2013 #1
    Hi,

    I'm sure this topic has come up before, however, it was always under the requirements of being realistic, i.e. taking drag, air resistance, earth rotation and what not into account. Also, I couldn't find any equations so that I can do the math myself. Hence, here is my question:

    Suppose I shoot a bullet at 400m/s vertically into the air, I keep reading that it would arrive on the ground at the same speed. Now, I am a really not good at math, but I "know" that under the gravitational force of the earth, any object falls (counting air density, resistance, rotation, etc. out) at the speed of 9,81m/s.

    I find it hard to imagine that, knowing this, a bullet would accelerate to its initial speed. So, I wanted to calculate it myself. Here comes the math:

    I figured, if the bullet has a speed of 400m/s, then gravity would stop the bullet after aprx. 8154,94m (h = u² / 2g). Knowing its speed, that would mean it would take the bullet 20,39 seconds.

    If the bullet arrives at the ground with the same speed, shouldn't it also arrive within the same time frame? I calculated that it will actually take the bullet 40,77 seconds to reach ground (t = √(2d/g)).

    That means that the speed of the bullet is actually 199,98 m/s (v = ½ ⁎ g ⁎ d), so half as fast.

    m=meters
    s=seconds
    h=height
    u=initial velocity
    g=gravitation
    d=distance
    v=velocity

    Now, I am sure I made a super obvious mistake here, probably using the wrong equation or a simple, logical mistake.

    Would anybody please be so kind and point it out to me and, if possible, provide with a correct trail of though/math so that I don't have to send the university of illinois physics department an email? :)

    Thanx so much!
     
  2. jcsd
  3. Dec 9, 2013 #2

    Nugatory

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    The time isn't right, although the height is. If the bullet starts out at 400 m/sec, and gravity is slowing it by 9.8 m/sec every second, it stops after about 40 seconds at a height of about 8000 meters. I suspect that you forgot that the average speed of the bullet is 200 m/sec, not 400 m/sec.

    Now you have the time right, but your calculation of the speed is wrong. Instead, you can use the much simpler ##v=at## which relates speed to time spent accelerating.
     
    Last edited: Dec 9, 2013
  4. Dec 9, 2013 #3

    D H

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    Welcome to PhysicsForums, Sairen!

    You made two mistakes in your calculations. Here's the first one:
    You didn't show us how you arrived at that figure, but it's wrong. It takes the bullet 40.77 seconds to reach that maximum altitude, the same amount of time as it takes to fall from that altitude.

    And that's the other. You apparently used ##v=\sqrt{gd/2}##. You should have used ##v=\sqrt{2gd}##.
     
  5. Dec 9, 2013 #4
    This is not correct. 9.81 is not a speed in metres per second, it is the acceleration due to gravity i.e. the rate of change in speed - each second the speed of a free falling object changes by 9.81 meters per second, so we say the acceleration is 9.81 metres per second per second or 9.81 ms-2 (or m/s2).

    This is correct, although I wonder where you got the formula h = u² / 2g from?

    You have assumed the bullet is travelling at a constant speed of 400 m/s, but we know that the speed is not constant because when the bullet stops its speed is 0. In fact the speed decreases at a constant rate (of 9.81 ms-2) so its average speed is 200 m/s. It therefore takes the bullet twice the time you have calculated to travel this distance.

    This is correct, and is the same as the correct ascent time.

    Again this is the average speed; the final speed is twice this i.e. 400 m/s.
     
  6. Dec 9, 2013 #5

    SteamKing

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    This is one of the things you 'know' that just ain't so:

    Objects falling under the influence of gravity don't travel at a speed of 9.81 m/s. They accelerate at 9.81 m/s^2, which means if you drop something, after the first second, it reaches a velocity of v = at or 9.81 m/s, after two seconds, v = 19.62 m/s, and so on, neglecting air drag, etc.
     
  7. Dec 9, 2013 #6
    Nugatory, totally! Didn't see that..


    Thanks D H! As for the calculation: I simply calculated 8154,94 / 400 = 20,39.


    MrAnchovy, I got the formula from here :shy:


    SteamKing, I thought so. Hence, I put my "knowledge" in quotation marks :thumbs:



    Thanks to all of you for the swift answers! It's all clear now :!!)
     
    Last edited: Dec 9, 2013
  8. Dec 9, 2013 #7

    K^2

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    This only works if the average speed was 400m/s. (Or if it was always 400m/s, which amounts to the same thing.) The bullet, however, slows down on the way up, and in fact, the average speed is going to be 200m/s. (This is true for any constant deceleration.) So time in flight is going to be twice what you computed, matching time you (correctly) computed for the bullet falling down.

    Of course, all of this assumes no air resistance. Real bullets never fall at the speed they were fired due to air resistance. So it's not something you can test with a bullet. But a steel bearing thrown or fired out of a small spring-loaded cannon vertically up is affected by air resistance very little, so you can test it out and it does work. It falls down at the same speed that it was thrown/fired at.
     
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