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Simple question about gravity and force

  1. Oct 5, 2003 #1
    Simple question about gravity and force....

    I'm hoping you can help. My husband and I have been discussing a situation and we would love to find an answer. Any direction you can give me would be great. (Please note, I haven't taken a physics class in about 15 years)

    If you shot someone in the head with a gun at point blank range and you shot the same gun straight up into the air and the bullet came down and hit the person on the head would you be shooting this person with the same amount of force and/or speed?
     
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  3. Oct 5, 2003 #2

    LURCH

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    Except for the loss of force due to air resistance (which is actually quite a lot, since it resists both on the way up and on the way down), yes.

    NOTE: Please do not attempt to verify this experimentally!
     
  4. Oct 5, 2003 #3
    I would guess absolutely no. If you shot someone point blank it would kill them. But if you dropped a bullet from the sky and it hit someone's head it would hardly even hurt them.

    A gun creates a force to push a bullet. And if you shot it straight up the force would take it high in the sky. But the force would eventually weaken and loose all of the force created by the gun; the bullet would stop in mid air and then change direction. At this time another force acts on the bullet, gravity. Gravity will pull the bullet down and it will accelerate at 9.8 meters per second per second.

    I have no idea what the acceleration of a bullet shot from a gun would be, but I know it would be a lot more than the force of gravity. If this wasn’t the case we would hear horror stories all the time about injuries caused by bird crap hitting humans.

    This brings up another question though... If earths gravity accelerates free falling objects at 9.8 m/s/s, then how fast could that object go before it hit the earth?

    Is there a limit? The faster an object falls the greater its mass becomes, and the greater the mass the more force is needed to accelerate it. So is there a limit to how fast something can fall to earth? Wont the gravitational effects of earth not be enough of a force to push an object any faster than it is already moving once the object reaches a certain mass?
     
  5. Oct 5, 2003 #4

    Integral

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    The only force acting on the bullet after it leaves the muzzle it that of gravity. The bullet will accelerate at g, the entire time it is in the air. For the first half of the time g will serve to slow the bullet, until, as you said, it comes to a stop, from that point on it will increase its speed, still accelerating at g, until it strikes the ground traveling at THE SAME SPEED at which it was fired.

    As Lurch said there will be losses due to air resistance, so the bullet will be traveling slower then the point blank range firing. The real question is, what is the terminal velocity of the bullet. I would bet that it is quite high, high enough to be fatal it it struck someones head.
     
  6. Oct 5, 2003 #5
    I am a newb, and I don’t mean to argue, but I disagree the bullet would be traveling at the same speed as it was fired when the bullet is free falling; even if we forget about the air resistance.

    It would be no different than if you dropped the bullet from a hovering helicopter a few miles high in the sky.

    You said the only force acting on the bullet after it leaves the muzzle is G. I agree with that but when you say G accelerates the bullet when fired upward I disagree. G will decelerate the bullet until it stops and then G will pull it down at 9.8 m/s/s.

    What I am saying is the acceleration the gun gives the bullet when the bullet is fired will be much greater than 9.8 m/s/s. And the terminal velocity of a free falling bullet would be less than the velocity of a bullet fired from a gun at point blank range.

    I haven’t done this experiment but it is easy to imagine. We all know if we hold two different bullets with two different hands, three feet from the ground, and drop them at the same time they will land on the ground at the same time. But if one hand holds a bullet three feet from the ground and another hand holds a gun with a bullet in it three feet from the ground and you fire one bullet (downward) at the same time you drop the other bullet the bullet fired will hit the ground first.

    *please note when the word "downward" again. I understand if you fire a bullet straight ahead three feet above the ground and drop another bullet three feet above the ground then they will land at the same time*

    Bullets kill people because of the force a gun gives a bullet. Bullets do not kill people because of the gravitational force the earth gives the bullet.

    Again, if you drop a bullet from the sky it would have the same free falling speed when it falls as a bullet fired upward from a gun will have when it falls (provided the distance traveled once they begin to fall is the same for both bullets.) The gun and the explosion it creates to push the bullet upward is absolutely meaningless once the bullet stops and begins to fall.

    PS Thanks for clearing me up though on my 1st post. I agree totaly the only force that acts on a bullet is G after it is fired. I was wrong thinking a bullet accelerated what so ever after it leaves the muzzel. And I was wrong thinking G didnt act on the bullet untill it began to fall.
     
    Last edited: Oct 5, 2003
  7. Oct 5, 2003 #6

    chroot

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    syano,

    You don't know what the hell you're talking about, I'm sorry. A projectile fired upwards with velocity v will come down with the same velocity. This is due to the conservation of energy; gravity is a conservative force. There is no question about this fact. The bullet begins it journey upwards with some amount of kinetic energy; as it ascends this kinetic energy is traded for potential energy. At the point where the bullet stops, it has no kinetic energy, and its maximum amount of potential energy. As it falls, this potential energy is traded back for kinetic energy. The sum of kinetic + potential energy is the same at all times. No energy is lost. The bullet goes up with velocity v, and comes down (to the same height) with the same velocity, now pointed downwards.

    The problem you seem to be missing is that the acceleration due to gravity, g, affect the bullet for a very long time -- the entire time it's in the air, in fact. The acceleration due to the gun is obviously much larger, but it only affects the bullet for a very short time.

    Integral and Lurch are correct; you are not. Learn some basic mechanics before trying to argue with people. This is high school stuff.

    - Warren
     
  8. Oct 5, 2003 #7

    HallsofIvy

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    But you also said "even if we forget about the air resistance."
    If there is no air resistance then there is no "terminal velocity".

    If a bullet if fired directly upward and there is no air resistance, then you could calculate exactly the time and distance until its upward velocity is zero. It would then fall downward under exactly the same acceleration for exactly the same time and distance.
    When the bullet reaches its original level, it will then have exactly the same speed as when it was orinally fired but with opposite direction.
     
  9. Oct 5, 2003 #8

    Janus

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    The higher you drop the bullet from, the greater the velocity it will have when it hits.

    The velocity the gun leaves the bullet determines how high the bullet will be when it stops climbing and starts falling. Therefore the force of the explosion determines how high the bullet will travel, and how hard it will hit when it come down. For instance, if the bullet had a a muzzle velocity of 980m/s, then decelerating at 9.8m/s², it would take
    10 sec for the bullet to decelerate to zero and start falling. The height it would reach would be

    D = at²/2 = 9.8(10²)/2 = 490m

    The time it would take to fall from this height is

    t = [squ](2d/a) = [squ](2(490)/9.8) = 10 sec.

    The same amount of time it took to travel up to that distance.

    Accelerating at 9.8m/s, it would be traveling at 9.8*10 = 980 m/s when it hit. the exact same velocity it had when it left the gun.

    if you bump the muzzle velocity up to 1960 m/s, it would travel upwards for 20 seconds and gain an altitude of 1960 m. It would again take 20 sec to fall from that height, thereby regaining its lost velocity and reaching 1960 m/s on impact again.

    Terminal velocity takes air resistance into account. Terminal velecity is determined greatly by the denstiy of the object. (this is why people are not injured by falling bird poop; bird poop is much less dense than a bullet, and thus has a much lower terminal velocity in air. People and property have been injured or damaged by falling bullets, which is one reason city have laws against discharging weapons within city limits.

    But we are, for the moment, ignoring air resistance, thus there is no terminal velocity to consider.
     
  10. Oct 5, 2003 #9

    Integral

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    Then perhaps you shouldn't :wink:

    To a physicists the only difference between acceleration and deceleration is a negative sign, deceleration is simply negative acceleration, so the bullet accelerates at g the entire trip.

    Yes so the only important number is the terminal velocity of the bullet, it will decelerate to that speed and when it returns to earth it will be traveling at that speed.
    In this case you have different initial conditions, when the bullet reaches some height it has zero velocity, if I dropped a bullet from the same height with zero velocity it would be traveling with the same velocity when it reaches the ground as the one fired up wards from the gun. It does not matter how the bullet got to the height only that it is there and has zero speed.

    As I stated before the critical number is the terminal velocity, I still believe that it would be fatal.

    If you ignore air resistance then there is no terminal velocity, the bullet will return to the earth with its original speed. Think about it. It is decelerated (negative acceleration) by g to some height then returns the same distance being accelerated by g, how can the return velocity be different?
     
  11. Oct 5, 2003 #10

    Nereid

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    Suggestion

    For syano:

    You can test this for yourself, with a simple experiment that doesn't involve bullets.

    You and a friend find a tree, or an overhanging second storey balcony (or first storey, depends on which part of the world you're in). Get a couple of marbles. Have your friend climb to a height about as high as you can throw a marble. You throw one marble up, to the height your friend is, and catch it when it comes down (I hope you can't throw too hard; it'll hurt). Your friend drops the other marble and you catch it.

    Which marble hurt the most when you caught it?

    'Hand-hurting' isn't a very accurate way to determine the kinetic energy of the marbles; perhaps you can think of a better way?
     
  12. Oct 5, 2003 #11
    Thanks for the explanation HallsofIvy, Janus, Integral, and Nereid!

    I am beginning to understand :-)

    Nereid’s experiment really throws me off. Nereid in your experiment would the marble thrown up and then caught hurt more than the one that was dropped? If so that is really neat.

    If I threw a marble straight up, and it reached 20 meters before it fell, and I caught it in my hand…would that hurt more than if a person was standing on a building 20 meters high and dropped a marble and I caught it?

    If the thrown marble would hurt more than the dropped marble, than my analogy of dropping a bullet from a hovering helicopter was completely wrong.

    This is intriguing!!!
     
  13. Oct 5, 2003 #12
    The marble dropped from 20 meters will hurt more so Nereid's experiment doesn't help clarify.

    Instead, think of conservation of
    energy: -mgh = mgh so initial and terminal velocity has to be the same. If you decelerated constantly going at 10m/sec for 100 meters in a car to reach 0 velocity and wanted to know what velocity your car will reach going from 0, accelerating at the same rate for 100 meters, what will your answer be? So, a bullet shot at point blank will have the same terminal velocity as one that was shot upwards (
    neglecting air resistance).
     
    Last edited by a moderator: Oct 5, 2003
  14. Oct 5, 2003 #13

    Integral

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    Why? Hasn't been well established in this thread that it will have the same velocity, therefore the same energy if it thrown up a height,h, or dropped from the same distance h. Wouldn't they feel the same?
     
  15. Oct 5, 2003 #14
    A marble thrown upwards loses kinetic energy while a marble dropped
    downwards gains kinetic energy. More kinetic energy = higher velocity = more momentum= hurts more.
     
  16. Oct 5, 2003 #15

    Janus

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    Um, the marble being thrown up falls back , and gains back the kinetic energy it lost going up. The marble is caught by the person on the ground who threw it up in the first place,

    As long as the dropped marble is dropped from the same height the thrown marble reaches at the peak of its trajectory, they both will have the same momentum when caught.
     
  17. Oct 5, 2003 #16

    Integral

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    Yes the thrown marble will lose kinetic energy, but not total energy, the kinetic energy is converted to potential energy. When it reaches the top of its motion, it will turn, fall back to the ground and regain ALL of its kinetic energy, striking the ground at the same speed at which it was thrown.

    Kindly take some time and read the other posts in this thread.
     
  18. Oct 5, 2003 #17
    Nevermind, I misread Nereid's experiment. I thought his friend
    was supposed to catch the marble on its way up.
     
  19. Oct 5, 2003 #18
    Re: Suggestion

    How about letting the marbles drop into a uniform layer of mud or clay and measuring the penetration depth? I believe that's the way they did it in Galileo's day.

    However, the distance into the clay will not be directly proportional to the velocity; it will be proportional to the square of the velocity.

    In terms of pain, a marble going twice as fast creates four times the pain...He, he.:wink:

    Creator
     
    Last edited: Oct 5, 2003
  20. Oct 5, 2003 #19
    Musket balls were once produced by dropping a measure of molten lead from a tower made especially for that purpose. By the time the projectile reached water below, it was relativity solid and spherical.

    You all have heard not to drop pennies off of the Empire State building? Assuming the ESB 1000 feet tall, and the penny having no frictional forces when dropped with sides horizontal, the penny's velocity is given by v=(2gH)1/2=(2x32x1000)1/2~250 ft/sec. A 22-short bullet has a fired velocity of ~500 ft/sec.

    For forensic identification, bullets are often fired into a container (the size and contents of a water barrel). Likewise, jello can be used to simulate human tissue. Either water or colloid could be used to determine the kinetic energy of a falling object.
     
  21. Oct 6, 2003 #20
    I cannot beleive the amount of debate here over a simple question. Some people need to go and read a book on mechanics, and when trying to explain something like this, NEVER neglect air resistance. Fine conceptually, but a very unrealistic model on our world.

    Jonathan
     
  22. Oct 6, 2003 #21

    LURCH

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    Ah good; I was going to make sure to visit this thread today to make exactly that same point (about air resistance). Although the title of the thread and the original question led me to believe that the original poster was only interested in the effect gravity would have on the force of the bullet. Even so, in my initial post I did feel compelled to at least mention the fact that air resistance would make a large difference. In fact, air resistance is the only difference between being shot at close range and being shot from a distance.
     
  23. Oct 6, 2003 #22

    Nereid

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    as an experiment in school?

    There are images you see from time to time on the TV news of people in a crowd firing into the air, AK-47's maybe. I remember reading once that in such situations there are often people injured by the falling bullets, and sometimes killed. I did a quick google, but couldn't find anything. It surely happens; does anyone have a specific reference to a report of injuries or deaths so caused?

    Some of the discussion in this thread suggests that it's not an easy concept to take in, so I was wondering if a simple experiment could be built, for a high school physics class perhaps. Of course, guns and bullets would NOT be a good idea! However, perhaps the spring etc from a pin-ball machine? And I like the soft clay as a means of measuring kinetic energy. Most likely a teacher with energy and time (pretty rare these days I'm sure) putting one together by herself; getting the 3 R's right may not be easy (reliability, repeatability, robustness).
     
  24. Oct 6, 2003 #23
    In a previous thread, it was stated that all objects (regardless of their mass) free-fall with the same acceleration - 9.8 m/s/s. This particular acceleration value is so important in physics that it has its own peculiar name - the acceleration of gravity - and its own peculiar symbol - "g" not capital <B>G<?B> (thats Universal). But why do all objects free-fall at the same rate of acceleration regardless of their mass? Is it because they all weigh the same? ... because they all have the same gravity? ... because the air resistance is the same for each?

    In addition to an exploration of free-fall, the motion of objects which encounter air resistance will also be analyzed. In particular, two questions will be explored:

    Why do objects which encounter air resistance ultimately reach a terminal velocity?
    In situations in which there is air resistance, why do more massive objects fall faster than less massive objects?
    To answer the above questions, Newton's second law of motion (F net = m*a) will be applied to analyze the motion of object which are falling under the sole influence of gravity (free-fall) and under the dual influence of gravity and air resistance.

    Free-fall is a special type of motion in which the only force acting upon an object is gravity. Objects which are said to be undergoing free-fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass. But why? Consider the free-falling motion of a 10-kg rock and a 1-kg rock.

    Bigrock -> m=10kg -->Smallrock is m=1 kg
    Fgrav for Br=100 N using Newtons second law the formula is used: a= F/m --> when doing this you plug in the units. A= 100 N/10 kg; a=10 m/s^2. Do the same for the Smallrock.

    So here: The mass for small-rock is 1 k-g. So the Fgravity for the Smallrock=10N --> Using the same equation the formula will look like this: a=10N/1 k-g --- then a=10 m/s^2.

    TADA! You get the same acceleration! Isn't that insane? No, it's not.
    If Newton's second law were applied to their falling motion, and if a free-body diagram were constructed, then it would be seen that the 10-kg rock would experiences a greater force of gravity. This greater force of gravity would have a direct effect upon the rock's acceleration; thus, based on force alone, it might be thought that the 10-kg rock would accelerate faster. But acceleration depends upon two factors: force and mass. The 10-kg rock obviously has more mass (or inertia). This increased mass has an inverse effect upon the rock's acceleration. And thus, the direct effect of greater force on the 10-kg rock is offset by the inverse effect of the greater mass of the 10-kg mass; and so each rock accelerates at the same rate - 10 m/s/s. The ratio of force to mass (Fnet/m) is the same for each rock under situations involving free fall; this ratio (Fnet/m) is equivalent to the acceleration of the object. Gottit? Okay...

    Integral did a nice job of explaining it to you, but you didn't understand, syano, listen to the man -- HE KNOWS WHAT HE'S TALKING ABOUT!
     
  25. Oct 6, 2003 #24

    krab

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    Re: as an experiment in school?

    Try here

    Also, in Google, try

    "terminal velocity" bullet

    Lotsa stuff. One such page is http://www.loadammo.com/Topics/March01.htm [Broken]. They claim that it requires 60 ft-lbs. for a disabling wound, and a falling 30 cal bullet has terminal velocity such that its energy is 30 ft-lbs. So, not so bad. But consider this. Imagine someone drops a 30 lb. weight on your head from a height of 1 foot. Still maybe survivable. Now imagine the weight has a point pointing downward, whose cross section is the same as that of a 30 cal bullet. Ouch.
     
    Last edited by a moderator: May 1, 2017
  26. Oct 6, 2003 #25
    Thanks Jeebus! Yes you are right. Integral certainly did an excellent job of explaining this to me, as well as Warren/Chroot. I didn’t thank Warren in my last post because I was a tad upset with how he addressed me. No biggie though, cause it really started to sink in with me when he explained how the bullet starts with some amount of kinetic energy that is traded for potential energy as it moves upward, and at its stopping point in mid air it has its maximum amount of potential energy and no amount of kinetic energy.

    And then he really made it sink in and stick with me, when he said that I was forgetting the acceleration of the bullet fired from the gun only last while the bullet is in the muzzle, but the acceleration of gravity will act on the object through its entire journey.

    So, Thanks Warren :-)

    Jeebus I knew where you were going when you were talking about why 2 objects of different mass fall at the same rate. I am on chapter 3 of Hawking’s Brief History of Time, and I recalled him saying “a body of twice the weight will have twice the force of gravity pulling it down, but it will also have twice the mass. According to Newton’s second law, these two effects will exactly cancel each other out, so the acceleration will be the same in all cases”

    Oh, and thanks for clearing me up on the difference between “G” and “g” as well Jeebus.

    ** Syano, holds his head up high like Barney Fife from the Andy Griffin Show and says, “Hope I answered your question Oliva”**
     
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