HSC PHYSICS - HARD Projectile Motion Questions

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  • Опубликовано: 3 ноя 2024

Комментарии • 9

  • @mastermind23109
    @mastermind23109 Месяц назад +2

    Hi Ryan,
    Regarding Q22) HSC 2010 a) and b), shouldnt t=0, because you are trying to find the initial velocity, and the question gives you the time of flight of the whole duration of the rock returning to the ground at 21s. Thanks

  • @fatimasaad8678
    @fatimasaad8678 4 года назад +5

    This is amazing!

  • @bentuu816
    @bentuu816 11 месяцев назад +1

    what app are you using to work on these questions? (as in the overlay you're drawing on it with)

  • @1Happy_Singh
    @1Happy_Singh 2 года назад

    1st Answer Seems Wrong Because The Velocity Is
    V(final)= 1.5 x sin(50 degree)
    Can You Please Explain Me If I Am Wrong...
    Please

    • @RyanLasek
      @RyanLasek  2 года назад +1

      By "1st Answer", I'm guessing you're refering to HSC 2019 Q30 part a.
      The magnitude of V(final) is 1.5 m/s. It's direction is 50 degrees below the horizontal.
      You've written that: "V(final)= 1.5 x sin(50 degree)" which is wrong.
      "1.5 x sin(50 degree)" is equal to the *vertical component* of the final velocity.
      Here are the official answers. Check question 30: educationstandards.nsw.edu.au/wps/wcm/connect/b24c0c95-329e-4f2b-94dc-c8e588dd942d/2019-hsc-physics-mg.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-b24c0c95-329e-4f2b-94dc-c8e588dd942d-no2b2Dl

    • @1Happy_Singh
      @1Happy_Singh 2 года назад +1

      @@RyanLasek Thank You ! 😃

  • @wildepicmovies1044
    @wildepicmovies1044 2 года назад

    in question one you have canceled both g out from the left side but why did you add one to the right

    • @RyanLasek
      @RyanLasek  2 года назад

      I think you're refering to question 1 part a at 5:03.
      In this question, I'm trying to find the change in height.
      In order to find the change in height, I have to get a (hf - hi) term on it's own (because final height minus initial height is the change in height I'm looking for).
      In order to get a (hf - hi) term on it's own, I need to simplify the equation and re-arrange it.
      I "cancelled out" the g's on the left side of the equation by *dividing* the left side by g.
      If I divide the left side of an equation by g, I need to divide the right side by g as well to keep the equation balanced.
      When I divide the right side by g, there were be a g in the denominataor.
      Say you had an equation like 2x = 8 and you wanted to find x.
      In order to find x, you need to divide the left side by 2. Then, the to keep the equation balanced, you also need to divide the right side by 2.
      You'll be left with: x = 8 / 2.
      The 2 has been "cancelled out" from the left side, but now there is a 2 in the denominator on the right side.