Projectile Motion in Sports

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  • Опубліковано 15 жов 2024

КОМЕНТАРІ • 113

  • @alireza98325
    @alireza98325 3 роки тому +18

    The 5.80 m is not the distance from the foul line to the center of the basket.
    It should be 4.191m = 13.75 feet
    The foul line distance is 15 feet from the front of the backboard.
    The distance from the center of the basket and the front of the backboard is 1 foot 3 inches (1.25').
    Thus the distance from the foul line to the center of the basket is 15'-1.25'=13.75'=4.191meters

  • @jefferywilliams6984
    @jefferywilliams6984 4 роки тому +13

    G.O.A.T.!!!! You helped me with my physics project!

  • @awesomeguy1303
    @awesomeguy1303 5 років тому +18

    Really helpful, one problem tho dwight is crap at free throws

    • @appurajosh
      @appurajosh 3 роки тому

      Dwight as in Dwight Schrute?

  • @hoangkhai2377
    @hoangkhai2377 8 років тому +8

    This video theory-wise is really good but there is indeed a flaw in the working.
    The final velocity of x and y of the shot should not be 0 considering that the ball would still be moving even though it has landed inside the hoop. I have been analyzing a lot a free throw videos for my essay "model of a perfect free throw" and found that the average landing velocity of a basketball for x would be: 3.53m/s and y to be -3.32m/s. Just plug those numbers into the original set of equations and you should be able to figure out the correct Velocity and initial angle of release.
    This is because it would be nearly impossible to make a free throw at an angle of 30.8 because the angular diameter of the hoop would be so small, the ball wouldn't able to fit.

    • @takoalawson5165
      @takoalawson5165 8 років тому +2

      This video was completed by a student taking Regents Physics and the generalizations are to the spirit of the level of that course. But as you noted it is not a perfect model for an actual free throw. The math can be worked out just as easily but the fact the hoop is not the same altitude as the launch point and as you noted the ball is actually returning towards earth as it passes through the net (not apex or vertical velocity of zero) are beyond the scope of what the regents curriculum covers so we're not included in the concepts and skills discussed by the creator during production. I think there are some physics based sports shows that have a more accurate model/description of free throws actually taken in games, which you have probably come across doing your research. Great eye anyway, it's good to be mindful of these types of details when trying to build an accurate physical understanding of our universe.

    • @hoangkhai2377
      @hoangkhai2377 8 років тому +1

      Haha I still really enjoyed this video by the way :) If you have a chance to meet the student who made this- please tell him his 3D animation and stuff look really cool

    • @takoalawson5165
      @takoalawson5165 8 років тому +4

      +Hoang Khai I am glad you enjoy it. I also think it is a very well done and impressive piece of work. The production is one of the best I have had a student submit. I had the great fortune of having the creator as a student for a year and still get in contact with him now and then. A very bright and intelligent young man who has great balance between art and analytics. I am confident his work ethic and intelligence will bring him great success in life.

    • @ekinozsoy8017
      @ekinozsoy8017 4 роки тому +1

      @@takoalawson5165 Your student must be so lucky to have a supportive teacher like you. I hope he read all the compliments that you wrote for him.

    • @DigitalLawson
      @DigitalLawson  4 роки тому +1

      Ekin Özsoy thank you for the kind words. This particular student and I have had communication after graduation since he was considering a career in education, from our last conversation I believe he is even considering science and teaching physics specifically. It is fun imagining him having physics students who find a video he made about the topic when he was in high school. Cheers to all students and the continued success of their exploration through education.

  • @jeehadmuhammad1534
    @jeehadmuhammad1534 3 роки тому +2

    the ring is not on the baseline. the distance between the baseline and the ring is 1.575 m. so distance between dwight howard and the ring is 5.80 - 1.575 = 4.225 m. but the video is already great

  • @jakobtamas5885
    @jakobtamas5885 8 років тому +3

    I was told that "S = UT + 1/2 AT(squared)" was used to find the vertical values, not the horizontal value. Horizontal I thought was "S = UT".

    • @takoalawson5165
      @takoalawson5165 8 років тому +1

      Since the horizontal acceleration is zero, S = UT + 1/2AT(squared) becomes S = UT! Anything times zero is zero, just depends if you want to ink use the fact the acceleration is zero in your work, which this student did when they created this project. Hope this helps.

    • @takoalawson5165
      @takoalawson5165 8 років тому

      +Takoa Lawson *include not ink use

  • @shanchuanyou2477
    @shanchuanyou2477 3 роки тому +1

    Thank you! This helps me a lot since I am thinking modeling a free kick in soccer match which may have similar situation!

  • @marissawhitehead5670
    @marissawhitehead5670 3 роки тому +2

    For the first eqn, Im trying to find vi, but Im not sure how. I watched other videos and it said to set it up as vi = square root of (vf2-2ad) but every time I get a negative number because vf2 equals 0. Could you help me?

  • @instudio2408
    @instudio2408 3 роки тому +1

    what apps do you use to make the animation? 😍😍 really cool!

  • @samyuktha6218
    @samyuktha6218 5 місяців тому

    Why vf = 0? Since gravity acts in vacuum as well

  • @sabrysm
    @sabrysm 3 роки тому +1

    But how do I manage my hand to throw the ball with 10 m/s how fast should i move my hand ? And for making such an angle, what is my x-coordinate will be ?

    • @DigitalLawson
      @DigitalLawson  3 роки тому +1

      That’s more a question for a shooting robot, a person can rarely achieve an exact speed or angle fit a throw and while some video analysis can help you make better shots it’s good to just start practicing shooting and build a muscle memory for the force and angle that gets the ball where you want it.

  • @albertperez2502
    @albertperez2502 5 місяців тому

    How did you get 13.18 for vix?

  • @Mar1-YT
    @Mar1-YT 3 роки тому +1

    hi sorry for bothering you but when i put the values to find time i kept getting 5.52, how did you get 0.44?????

    • @DigitalLawson
      @DigitalLawson  3 роки тому +1

      Time equals -4.29m/s divided by -9.81m/s^2, you subtracted them which is an algebra rules error.

    • @Mar1-YT
      @Mar1-YT 3 роки тому

      @@DigitalLawson Thank you very much sir, you have helped me soo much you don't even understand.

  • @CssGamer28
    @CssGamer28 7 років тому +2

    I can't see any reason why you use the distance from the free throw line to the backline in the equation. I would argue that you should use the the distance from the free throw line to directly under the middle of the basket. Otherwise you would get some extra length.

    • @DigitalLawson
      @DigitalLawson  7 років тому

      Carl Johansen I would suggest you read previous comments particularly the one noting "it's not a perfect model" the narrator does state from the basket to the free throw line, the animation appears to extend a little further, but that could just be parallax error because of the vantage point of the camera. In either case it's a reasonable approximation for the course and demonstrates some concepts related to projectile motion.

    • @Shubh2004-shubh
      @Shubh2004-shubh 6 років тому

      nice

    • @Shubh2004-shubh
      @Shubh2004-shubh 6 років тому

      but how did you find the angle?

  • @vishalkumaryadav5534
    @vishalkumaryadav5534 3 роки тому +1

    In indian it is 9th standard physics
    We can do it with projectile equation also

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      If your referring to the range equation that only works when the vertical displacement is zero. Since the basket and the launch point are at different elevations the range equation does not work for this type of problem. Similar to being launched of a cliff or onto a cliff at different elevations. This students solution has several simplifications that make it an imperfect description but not using the range equation is not one of them. If your just referring to the kinematic projectile motion equations generally then those are the ones that are used, although the variables may be slightly different depending on the teacher/country.

  • @MasterEngineerPh
    @MasterEngineerPh 3 роки тому +2

    This formula has alot of flaws, firstly if the velocity of the ball is 0 then it isn't a final velocity because it is still in it's max height. Final velocity should be in negative. Secondly, when shooting a ball, the arm extends which gives additional height in the y direction.

    • @DigitalLawson
      @DigitalLawson  3 роки тому +2

      The formulas do not have any flaws and match current kinematic formulas for constant acceleration. The velocity in the y direction is zero, I believe the student made this specification in the described solution. In this description the claim is that the maximum height occurs at the rim. It is a simplified explanation for a first year physics course and not intended to fully model the actual shot, not because it can’t be done but because the course this student took didn’t require those types of models be explained. A better model would treat the trajectory as an arc and have the maximum height between the shooter and the hoop but this explanation does not apply that detail.

  • @pengyutan4529
    @pengyutan4529 7 років тому +1

    i really enjoyed the video but why did u use the displacement between the hoop and player as the displacement when the final velocity is 0 instead of using the maximum vertical displacement when the ball is at the highest point?

  • @searchlockin951
    @searchlockin951 3 роки тому

    How did you get solve for theta

  • @omarnadeem369
    @omarnadeem369 5 років тому

    isnt it that for the initial vertical formula, the value used should be displacement? Using pythagoras theorem of the height and length?

    • @takoalawson5165
      @takoalawson5165 5 років тому +1

      Omar Nadeem the hypotenuse your describing would be the actual displacement of the ball from its initial position to the basket but when describing the parabolic motion of a projectile the strategy is to separate the vertical and horizontal descriptions of motion and work with them separately. That is why the student used the x and y displacements separately and didn’t use the hypotenuse your describing.

  • @abbysplaying6279
    @abbysplaying6279 11 місяців тому

    Where did -9.81m/s² came from?

  • @mikras9344
    @mikras9344 4 роки тому +1

    initial velocity can't be 13.8 m/s because you plugged the wrong time value 0.44s...the ball must do all the trajectory and then find the time for horizontal velocity which is bigger than 0.44s value

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      I’m not exactly sure what your referring to but for the video explanation given the time should be 0.44 sec. It is a simplified description of the shot and not actually what a basketball player would do because you would want the apex to be between the basket and the player but for this explanation the student made the basket the apex and for the high school curriculum it was related to that is an appropriate simplification.

    • @mikras9344
      @mikras9344 4 роки тому

      @@DigitalLawson so you say that you took the distance player - apex (which apex is the rim), as it was the full trajectory of the ball right?and therefore you plugged the time value related this trajectory..because of simplification of the problem,right?

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      I didn’t do anything, this was a student in my class that created an example scenario for this type of problem. He describes in detail what he did as part of his explaining the process in the video. You might review some of the previous comments related to the content and see if they help? Or clarify what the question is and I can try and help?

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      I did look at the video again and if you use the altitude difference between the rim and the player and the acceleration due to gravity and the vertical velocity of zero at the rim the time of flight is 0.437768 sec which is appropriately rounded to 0.44 sec. I am still not seeing what the confusion is?

    • @mikras9344
      @mikras9344 4 роки тому

      @@DigitalLawson The apex is when the ball gets to the rim or in the middle of player and the rim? Thats my question.. thank you

  • @othmanalhammad327
    @othmanalhammad327 3 роки тому

    Can someone tell me the website that you can calculate that equation?

  • @dimplejanepiala2332
    @dimplejanepiala2332 4 роки тому +1

    Can I ask? Where can you get the final velocity 0?

    • @DigitalLawson
      @DigitalLawson  4 роки тому +3

      When an object reaches the top of its trajectory the vertical velocity is zero.

    • @jonathanma3854
      @jonathanma3854 3 роки тому +2

      @@DigitalLawson But when the ball goes into the basket, the ball is not at the highest point, it has a downward velocity

  • @gwapoali1766
    @gwapoali1766 3 роки тому

    Can you tell me what is the importance of projectile motion in basketball?

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      Are you asking how a game that involves launching a ball and getting it to land in a very specific location might be related to a topic in physics that describes how objects that are launched can be modeled and predict exactly where they will land, how fast they are moving, the time it takes to get there, along with other relevant descriptions of motion? If that is your question I would say basketball relies on projectile motion for a team to earn points and projectile motion is a model of how the ball behaves when launched towards the rim.

  • @TREmreprogaming
    @TREmreprogaming 3 роки тому +1

    This shot wouldnt hit because you said the final velocity is 0 which means the ball must come from inside the cup not above and what about air drag

    • @teshubdinesh4159
      @teshubdinesh4159 3 місяці тому

      exactly, that was what i was just thinking

    • @TREmreprogaming
      @TREmreprogaming 3 місяці тому

      @@teshubdinesh4159 Why did you summon me here bro

  • @reenzpaoloa.lucero1070
    @reenzpaoloa.lucero1070 6 років тому +2

    How did you find the angle?

    • @DigitalLawson
      @DigitalLawson  6 років тому

      Reenz Lucero I think the student just used a rough approximation. There are several simplifications in the video and I’m not sure if he measured or just guessed the launch angle.

    • @Freshin000
      @Freshin000 5 років тому +1

      you could find the angle by using the horizontal initial velocity and the vertical initial velocity. you can use tangent to find the angle which is tangent^-1( y initial velocity/ x initial velocity) = tangent^-1 (4.29/13.18)=18.02971507

    • @Freshin000
      @Freshin000 5 років тому +1

      you could also use the cosine. cos^-1(13.18/13.86)=18.0219706

  • @evanbentedos
    @evanbentedos 3 роки тому +1

    how did you get 4.29?

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      The student shows the formula and values that were used to calculate the initial vertical velocity of 4.29 m/s in the video.

    • @evanbentedos
      @evanbentedos 3 роки тому

      @@DigitalLawson do you have a video on how to solve with that formula? because the answer was just shown, it wasn't explained or anything

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      If you mean how to do the algebra to isolate a variable there are lots of videos on the topic. It’s the only kinematic equation that doesn’t require time, if you do a Google or UA-cam search for how to solve the kinematic equation that’s doesn’t have time in it you should find lots of resources that go over the order of operations and how to utilize algebraic tools. The steps are exactly the same as solving any other algebra problem just the variables and values are specific to this topic in physics. The student chose to not extend the time further by showing each step of the algebra.

  • @dimplejanepiala2332
    @dimplejanepiala2332 4 роки тому

    And where did you get the time=0.44 and the angle=10.02°? Plsss answer me huhuhuhuhu

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      The video describes how those values are determined for the simplified model the student used.

  • @wkwkwkkwkwkwkkw9040
    @wkwkwkkwkwkwkkw9040 2 роки тому

    1:53 where did you get the acceleration

    • @DigitalLawson
      @DigitalLawson  2 роки тому

      When I’m free fall (ignoring air resistance) the horizontal and vertical acceleration are known constants.

  • @mubashshirahmad7299
    @mubashshirahmad7299 5 років тому

    What about air drag, friction from hand and internal pressure of ball?

    • @DigitalLawson
      @DigitalLawson  5 років тому +2

      This is a simplified description and air drag is negligible because of ratio of mass to surface area and max velocity reached. Friction from the hand is irrelevant because we are describing the motion as a projectile or while in the air and contact with the hand was prior to that motion. Although even if we considered friction it would only be relevant if we were interested in the force needed to cause the initial velocity and rotation of the ball during flight. The air pressure of the ball isn’t relevant as long as it’s sufficient to keep it a spherical shape, unless we were considering it making contact with the backboard, rim, floor, or hand. I am not sure why you are under the impression any of those factors determine the motion of a projectile while flying through the air?

  • @francisco_castro1859
    @francisco_castro1859 7 років тому +1

    isnt a=-9.81 why does he use 0 instead when doing horizontal velocity

    • @DigitalLawson
      @DigitalLawson  7 років тому +3

      Francisco_ Castro18 when describing projectile motion (2D) then its best to separate the motion into vertical and horizontal descriptions. The acceleration in the vertical direction when in free fall is due to gravity (g) which is 9.8 m/s^2 downward. However, in the absence of friction (air resistance) there is no net horizontal force, thus, the horizontal acceleration is zero. Let me know if that makes sense or if additional clarification is needed.

  • @thomasbataluna1087
    @thomasbataluna1087 4 роки тому

    Is it phytagoream theorem?

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      Thomas Bataluna is what Pythagorean’s theorem? There are several examples used in the video that involve right triangle mathematics. Some are related to the values of its sides (Pythagorean’s theorem) only but some also are relating to a specific angle and are specifically trigonometric functions.

  • @anilkumarsharma8901
    @anilkumarsharma8901 Рік тому

    compare it with standard parabola as well

  • @shivanandams9585
    @shivanandams9585 6 років тому

    what if a ball size is 2kg , how much force he need to apply

    • @takoalawson5165
      @takoalawson5165 6 років тому

      shivanandam s that is a different calculation and part of the motion. This video describes the motion of the ball after it has left the players hands. Your interested in calculating the average force applied by the player before the ball leaves the players hand and the information in this video would not apply. You would need information like the distance the player pushes the ball through making the shot or the time during which the hand is in contact with the ball before it leaves the hand for the shot.

    • @shivanandams9585
      @shivanandams9585 6 років тому

      hmm i want that calculation update me if u know that. scenario is 2 balls of different weight how much force is required to throw it. range is 20 meter elevation angle 45 degree

  • @francisco_castro1859
    @francisco_castro1859 7 років тому

    could u tell me the steps to find the horizontal velocity

    • @DigitalLawson
      @DigitalLawson  7 років тому

      Francisco_ Castro18 in the video the student is determining the vertical and horizontal velocities from the known displacement, acceleration, final velocity, and time. Once he determined the time from the vertical motion he was able to use the horizontal displacement and time to determine the horizontal velocity, since the horizontal acceleration is zero. In some projectile problems trigonometry would be used because the problem gives the initial velocity at an angle. However, in this example he is trying to determine the angle and initial velocity needed for the perfect free throw based on some simplifications and assumptions of the motion.

  • @thomasbataluna1087
    @thomasbataluna1087 4 роки тому

    Is it 9.81 or 9.18?

    • @DigitalLawson
      @DigitalLawson  4 роки тому +1

      Thomas Bataluna acceleration due to gravity (g = 9.81 m/s^2).

    • @thomasbataluna1087
      @thomasbataluna1087 4 роки тому

      Thank you so much you help us a lot

  • @annabellepyke
    @annabellepyke 7 років тому +1

    wow this really helps my tomahawk person.

  • @anzatzi
    @anzatzi 7 років тому +1

    nice, but you are calculating a brick shot off the front rim. imo--you need to allow for arc.

  • @georgemathew8218
    @georgemathew8218 4 роки тому

    Good Explaination. @

  • @bintangcahyapratama0640
    @bintangcahyapratama0640 4 роки тому

    Ok! Lets try it on the game :)

  • @surajitbosu8639
    @surajitbosu8639 6 років тому

    very nice video.

  • @antonjamescaballo1
    @antonjamescaballo1 3 роки тому

    Hello , how did you get 18.02?

  • @aadithyajohn4459
    @aadithyajohn4459 3 роки тому

    Please show steps of calculations i am 13 years old

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      The steps are shown in the video, there are free textbooks online that also break down the process if this students work is confusing for you.

  • @wkwkwkkwkwkwkkw9040
    @wkwkwkkwkwkwkkw9040 2 роки тому

    Can i know that topics

  • @kenkeanfoxie1635
    @kenkeanfoxie1635 4 роки тому

    Hi? may i ask? How did you get the 0.44 in time? im just confused

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      Which part of the video?

    • @DigitalLawson
      @DigitalLawson  4 роки тому

      By using the vertical motion described in the video.

  • @francisco_castro1859
    @francisco_castro1859 7 років тому

    okay i get i now

  • @omniballs9586
    @omniballs9586 Рік тому

    How to get the last one solution????

  • @chauzic7583
    @chauzic7583 7 років тому

    Now I see

  • @mubashshirahmad7299
    @mubashshirahmad7299 5 років тому

    Correct ideally but not in real world

    • @DigitalLawson
      @DigitalLawson  5 років тому

      Mubashshir Ahmad yes this is a high school Intro Physics class. Simplified models of projectile motion are usually a good foundation to have before looking at more detailed or complicated models. The most notable error in this motion is the students choice to have max height be the rim, in order to go through the hoop you need an arch so the max height would be in front of and above the rim somewhere depending on launch angle and velocity.

  • @chelseashinkfield4804
    @chelseashinkfield4804 7 років тому +1

    gate

  • @chelseashinkfield4804
    @chelseashinkfield4804 7 років тому

    **great

  • @lucaboles6336
    @lucaboles6336 3 роки тому

    Incorrect.

    • @DigitalLawson
      @DigitalLawson  3 роки тому

      Simplified model and incorrect have very different meanings.

  • @yauzi991
    @yauzi991 6 місяців тому

    i dont understand shit