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.: Physics in sport :.

  1. Jul 8, 2009 #1
    My problem
    Hello, i just started physics at school, and we had one week of it before the holidays. We got taught some basic things, VERY basic things such as what vectors and scalars are, VERY basic tables, and that's about it. And then the last day of the term the teacher tells us we have to do an assignment about physics in sport.

    Now because i was hardly taught anything in the first place, i seriously do not know the first thing about it.

    I've tried my best, and have spent over 3 hours researching physics in basketball, such as shooting a free throw and a vertical dunk, but cannot find much and can't do much either. It's an essay, so i won't post what I've done, (about 3 pages written :()

    So basically i just want to get some help regarding basic physics.

    I've found out that gravity attracts at around 32.2ft/s or so, so i tried to apply that to a basketball shot or a vertical jump, but didn't know how.
    2. Relevant equations
    I would really like to know how gravity affects an upwards motion. Example: For us to jump 1 foot in the air, what velocity would be required to do so? Initially i thought you have to make a force stronger than gravity.
    3. The attempt at a solution
    My attempt at a solution is 2 different 1/2 finished reports that probably are wrong, and if need be i will post them.

    Please help i'm so confused i've tried my best but what can you learn when you don't know what to study? :(
     
  2. jcsd
  3. Jul 8, 2009 #2

    LowlyPion

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    Welcome to PF.

    This link might be a useful place to start.
    http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l2b.html [Broken]

    There are other Lessons that that might help as well.
     
    Last edited by a moderator: May 4, 2017
  4. Jul 8, 2009 #3
    Pion thank you so much that site was very easy to understand and helped me a lot!

    But one thing i still don't 100% understand is, say if you jumped vertically. What is the velocity on average? and how does gravity reduce this?

    Does gravity reduce your jump velocity by 9.8m/s?

    Thanks for the site it was excellent!
     
  5. Jul 9, 2009 #4
    Gravity is a force that produces acceleration, not a one-time change in velocity. It doesn't slow you down by 9.8 m/s, it slows you down by 9.8 m/s per second.
    The units of gravitational acceleration are [tex]\frac{m}{s^2}[/tex]

    Average velocity is somewhat of an elusive term. What do you mean by that, exactly?
     
  6. Jul 9, 2009 #5
    I understand that it slows you by 9.8m/s, but what do you mean by m/s^2?

    Also, what i meant by average velocity, is the average velocity that people would create when they jumped vertically.

    And how "hard" would you have to jump to defy gravity and jump 1 foot off the ground?

    I really appreciate the answers guys keep them coming!
     
  7. Jul 9, 2009 #6
    Gravity doesn't slow down your velocity by 9.8 m/s.
    For every second you are under the effect of earth's gravity, you velocity towards the earth's center will be increased by 9.8 m/s.
    Your velocity is increased by 9.8 meters per second, per second. Hence the [tex]\frac{m}{s^2}[/tex] notation.

    The jumper doesn't quite create a velocity. He applies a force to the ground, the ground applies an equal and opposite force back on the jumper (This is an example of Newton's third law) which accelerates him until he loses contact with the ground, at which point he is no longer acted on by the ground's reaction force, and is subject solely to gravity.

    Once he's no longer in contact with the ground, he has a certain initial velocity.
    At that point, the only force acting on him is the force of gravity, which accelerates him towards the ground at a constant acceleration of 9.8 m/s² (I'll be rounding this to 10 m/s² for the purpose of this discussion, just so I don't have to reach for a calculator for my examples).
    That means that for every second he's not in contact with the ground, he will be 'falling faster' towards the ground.

    So if, for example he jumps hard enough to accelerate himself to an initial velocity of 15 m/s, it would take gravity 1.5 seconds to slow him down to rest. At this point he will be at his maximum height above the ground. Gravity would then 'slow' him down further.
    What goes up, must come down, or so the old adage goes. :)

    Putting all of this into mathematical form proves that the time it takes for the jumper to ascend is equal to the time it takes him to descend (I'm talking about the time when he's not in contact with the floor). You will also find that the magnitude of his velocity (Not its direction!) at every height is the same. That means that if he left the ground at 15 m/s pointing up, he would return to the ground at 15 m/s pointing down. If 0.5 seconds into the jump his velocity was 10 m/s, 0.5 seconds prior to hitting the ground, his velocity would be 10 m/s just as well.

    Let me ask you this, do you know some basic calculus (The concepts of what a derivate and integral are, and what the relation between the two is)?
     
  8. Jul 9, 2009 #7
    This was so helpful :D thanks so much! i'll try write my report now :D

    And sadly, no i do not know basic calculus, about the relationship.

    Thanks everyone for their replies i'll update soon :P
     
  9. Jul 9, 2009 #8
    Ok guys i'm doing well but one question:

    When you jump vertically, there are no horizontal forces working on you is there?
     
    Last edited: Jul 9, 2009
  10. Jul 10, 2009 #9
    Right, there are no horizontal forces
     
  11. Jul 11, 2009 #10
    ok thanks song and is it also correct to say that the jumper has to jump to a velocity of 33.2ft/s towards the ring to be 1 foot in the air at the time his velocity is 0, considering he is up for one second?
     
  12. Jul 11, 2009 #11

    Redbelly98

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    Hmm, no. If his velocity is 33.2 ft/s to the right, then he would be 33.2 feet to the right after 1 second.

    Or are you talking about 1 foot high off the ground?
     
  13. Jul 12, 2009 #12
    Yeah i meant off the ground, because wouldn't gravity make the difference of 1 foot?
     
  14. Jul 12, 2009 #13

    diazona

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    Actually, if a jumper jumped at 32 ft./s (which is pretty fast by human standards), he would be 16 ft. high, not just 1 ft., by the time his velocity was equal to zero.

    Here's an equation you can probably understand:
    [tex]v_f^2 = v_i^2 + 2 a d[/tex]
    This equation relates velocity at the beginning of some interval to velocity at the end of the interval. [itex]v_f[/itex] is the final velocity (at the end), [itex]v_i[/itex] is initial velocity (at the beginning), [itex]a[/itex] is the acceleration, and [itex]d[/itex] is the distance traveled. In your case I used [itex]a = -32 \mathrm{ft.}/\mathrm{s}[/itex], [itex]v_f = 0 \mathm{ft.}/\mathrm{s}[/itex], and [itex]v_i = 32 \mathrm{ft.}/\mathrm{s}[/itex] to come up with that 16 ft./s. You can't always get away with just multiplying the numbers you know :wink:

    Here's another equation for you:
    [tex]v_f = v_i + a t[/tex]
    This one also relates velocity at the beginning of an interval to velocity at the end of the interval, like the last equation, but instead of using the distance traveled, this one involves the time it took. So in your case, if you wanted to figure out how long it takes a jumper to reach his maximum height, you could use [itex]a = -32 \mathrm{ft.}/\mathrm{s}[/itex], [itex]v_f = 0 \mathm{ft.}/\mathrm{s}[/itex], and [itex]v_i = 32 \mathrm{ft.}/\mathrm{s}[/itex] as before, and you'd solve the equation to find [itex]t = 1\mathrm{s}[/itex]. That is, it takes 1 second for the jumper to get all the way up to his peak - and then it'll take him another whole second to fall down, for a total flight time of 2 seconds.

    And here's yet another one:
    [tex]d = v_i t + \frac{1}{2}a t^2[/tex]
    If you're clever you can derive this one from the other two.
     
  15. Jul 12, 2009 #14

    Redbelly98

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    Are you familiar with vectors? The answer depends on the jumper's vertical component of velocity when he leaves the ground. See the equations that diazona posted. The v's there are refering to the vertical component of velocity, which gravity affects.
     
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