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Projectile motion of tabletop ball?

  1. Feb 13, 2017 #1
    Hello,

    Could someone please check my answers to this physics problem? All my work and equations are clearly shown in the pictures below. If I got the problem wrong, please provide me with some suggestions as to how I can get it right . I'm not expecting anyone to just give me the answers :)

    Thanks!

    Problem 6

    You flick a ball from a tabletop, so that the ball lands on the floor 1.50 m below. The ball's initial speed is 2.10 m/s. Use g = 9.80 m/s2.

    (a) f the ball is launched off the table horizontally, what is the horizontal distance between the launch point and the point where the ball hits the floor?

    (b) If the ball is launched off the table at a 30° angle above the horizontal, what is the horizontal distance between the launch point and the point where the ball hits the floor?

    IMG_3082 2.JPG IMG_3083 2.JPG IMG_3084 2.JPG IMG_3085 2.JPG
     
  2. jcsd
  3. Feb 13, 2017 #2

    BvU

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    My compliments for your neat style of working !

    Don't you think it's strange the ball should travel so far in part a) ? Check your math for 2.10 * 0.31 = ....

    And if you find a flight time of 0.31 s in part a), wouldn't you expect it to be in the air a little longer in part b) ?

    The mistake here is that you let it land with a ##v_{f, y}=0##. Just before impact that is definitely not true !.
    You want to use one of the other SUVAT equations; can you determine which one you need ?
     
  4. Feb 13, 2017 #3

    BvU

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    A few more suggestions: In a), ##v_{\rm f, x} = 0 \ ## is not right: there is no force in the x-direction (you yourself write ##a=0## too), so the motion is uniform with constant speed in that direction: ##v_{\rm f, x} = v_{\rm i, x} = 2.10 \;##m/s.

    In b) you confuse yourself using ##v_{\rm f,x} ## for ##v_{\rm f,y} ##. And itis not zero.

    I have to admit to a too hasty response: I now see you split up the trip in two parts. One to the highest point and one from there to ground. Let me re-read and hopefully stand corrected ...
     
  5. Feb 13, 2017 #4

    BvU

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    Alas, I can follow the 2nd part of the trip to the point ##x_f = 1.5056 ## m (rounding off is not such a good idea here, I think... --- and I would use a notation that has only ##y## in the y direction, but never mind...).

    Then I see ##0 = ...## where you mix the initial ##y## at time t= 0.107 s and the initial ##v_{\rm i, y} ## at time t = 0. That's where things are going wrong. Fortunately, it's easy to fix: take both at t = 0 and solve the quadratic equation (you have to do that anyway) -- saves you half a page of extra work too !
     
  6. Feb 13, 2017 #5
    Hi BvU,

    thank you for your thorough explanation, and thanks for the compliment :wink:. Hopefully I interpreted your suggestions correctly when revising my work.

    For part a) I rechecked my multiplication and got and got 0.65 meters. Hopefully thats correct

    and I redid my work for part b) too (work posted below in the pictures). Would please mind checking again? Thanks

    IMG_3086.JPG IMG_3087.JPG IMG_3088.JPG
     
  7. Feb 13, 2017 #6

    gneill

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    For part (a) you wrote:
    upload_2017-2-13_13-55-49.png
    Check the calculation you did for the last line.
     
  8. Feb 13, 2017 #7

    BvU

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    Thanks @gneill , I sure missed that one !

    @Sagrebella:

    For part b) you now find the right answer, again in two steps: from 1.5 m to ##y_{\rm max}## and then from ##y_{\rm max}## to 0.
    Do you realize you can do it in one step by solving $$y(t) = 0 = y_0 + v_0\cos\left ({\pi\over 6}\right ) t - {1\over 2} g t^2 \quad \rm ?$$
     
  9. Feb 13, 2017 #8
    thank you for catching that error. There are so many equations and numbers its hard to keep track sometimes :P
     
  10. Feb 13, 2017 #9
    Thanks for checking my answer. And actually I was going to ask you about solving it in one step; I still don't quite understand. Would you mind elaborating a little more please?
     
  11. Feb 13, 2017 #10

    cnh1995

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    I believe there should be sin(π/6) instead of cos in BvU's equation for y(t).

    The y displacement of the ball is -1.5m when it hits the floor. Instead of writing it in two parts, you can write it in a single equation that BvU wrote. Take v=2.10sin30, a=-9.81 m/s2 and y= -1.5m.
     
  12. Feb 14, 2017 #11

    BvU

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    You bet there should ! Sorry for possible confusion !
     
  13. Feb 19, 2017 #12
    Hello,

    I know I asked this question a while ago, but I'm looking over my latest threads in order to review for a physics exam i'll be taking soon. I think I understand this problem, but would you mind explaining part b.) again please. How is it that I can solve for the total flight time of the ball using this one equation instead of breaking up the trip into two steps in the y-direction? Doesn't one have to consider the fact that the ball initially travels above the horizontal before landing on the ground?
     
  14. Feb 19, 2017 #13

    cnh1995

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    You have already considered that when you use vsin30° as the initial velocity and -g as the acceleration.
     
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