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Me, Thinking About A Jump and A Fall

  1. Sep 9, 2008 #1
    Hi all. I had grade 11 physics so I know the basics on physics.. grade 12 comes later this year... Anyways I was thinking about the systems in a jump, and the motion involved and I know motion physics work and all so don't get me wrong :P I'm not trying to disporve something here... I just know that I'm missing something in my thought of the systems in jump. Here's the situation I have theorized and the systems involved:

    Someone jumps up with a strong push from their "super" legs... That extra force from their legs gives them a high acceleration until the point that their feet leave the ground... at this point, they have to begin "decellerating" (I know.. acceleration in the negative direction) but they are decelerating from a high speed that their theoretical super legs gave them.. This will take some time, so that person reaches a "super jump" height of say 5 feet... Now, that person has the acceleration of gravity speeding them up for a distance of 5 feet...
    In my thought, I cannot see where gravity alone can make up the speed that the super legs gave the jumper. I mean, if gravity is constantly accelerating a person negatively, then I would think the jumper would go be going up faster (time-wise) than they fall down, so the sides (going up, vs going down) wouldn't be even... This cannot possibly happen though, right?

    Does anyone here see what I mean or am I not making any sense :P haha. I know I have typed a lot but I hope I made sense.

    Thanks in advance.
  2. jcsd
  3. Sep 9, 2008 #2


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    Staff: Mentor

    The acceleration and deceleration can certainly be uneven, though I'm not sure I see where your confusion is.
  4. Sep 9, 2008 #3
    Well, in a system such as this one (a jump), shouldn't the rise and fall time be equal? As dumb as this may sound, this question came up while I was watching Sports Science on Discovery the other day and they were saying "this jumper is following Newton's laws because his time from jump to peak, and from peak to ground were even" ... That's where I'm drawing my idea that the rise and fall should be equal. Maybe this is wrong?
  5. Sep 9, 2008 #4


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    Staff: Mentor

    Well, you need to be specific about what time periods you are talking about. The time the person is in the air is symmetrical, but that doesn't include the time he was pushing with his legs to accelerate. Once in the air, the influence of the legs is over and gravity is the only thing acting.
  6. Sep 9, 2008 #5
    ok well.. then here we go --> time in the air (good point by the way) . I see what you mean. Now, after the super jumper uses his super legs to accelerate to say... 30m/s^2, it would take 3 seconds (approx.) to reach the vertex of his jump right? So (since it has been over a full year since I've seen my uniform accel equations, although I do slightly remember them) would it take the same amount of time to fall ? approx 3 seconds?

    See, I think my mind is telling me that one side of the jump is unbalanced because I see the jumper's original push as giving him an advantage in speed over what gravity will be able to do in the same amount of time.... I know this must be wrong though... I just need it to be explained to me, and frankly you are doing a good job, and I thank you for it.

    Thanks !
  7. Sep 10, 2008 #6
    The jumper's speed off the ground is defined by his jump, a strong jump = fast rising. Maybe he is doing 20Km/hr upward the instant his feet leave the ground.

    Immediately he is in freefall, gravity is the only thing acting, he will reach the apex of his jump at a time given by the simple equation, and fall back. He will hit the ground at the same speed (reversed) that he left, i.e. 20Km/hr, air drag etc notwithstanding. The faster you jump the harder you fall.

    Fire a rifle straight upward on the moon, and you better not be standing under the bullet when it eventually comes back down!

    The time off the ground going up is the same as the time going down.

    Just go outside and throw a ball in the air if it doesn't make natural sense to you, throw the ball up slow then fast.
  8. Sep 10, 2008 #7
    wow, so that makes a lot of sense...

    What I understand then is that: the math works out so that you will rise a distance (d) based on your initial speed so that when you fall, gravity has enough distance (d) to speed you up to the same distance that you left.... If that makes sense.
  9. Sep 10, 2008 #8


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    Did you take into consideration that when the person hits the ground, they will have the same speed (downward) as they did initially (upward)? That would seem to me to show that the situation is NOT as "unbalanced" as you seem to think.
  10. Sep 10, 2008 #9
    yep, that makes sense . that's what I was saying in the last post. And so it makes sense.
    I enjoy these discussions :P Thanks for putting up with me everyone ! :D
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