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rock_mit182 Group Title

A 2.00-m-tall basketball player is standing on the floor 10.0 m from the basket, as in Figure P3.58. If he shoots the ball at a 40.0 angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m.

  • 6 months ago
  • 6 months ago

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  1. rock_mit182 Group Title
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    |dw:1397053258940:dw|

    • 6 months ago
  2. rock_mit182 Group Title
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    @skullpatrol

    • 6 months ago
  3. AkashdeepDeb Group Title
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    You need the equation of a parabola for this. Do you know what the parabola equation for projectile motion is? :)

    • 6 months ago
  4. rock_mit182 Group Title
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    \[y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]

    • 6 months ago
  5. rock_mit182 Group Title
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    that one ?

    • 6 months ago
  6. rock_mit182 Group Title
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    but i don't know about the intial velocity

    • 6 months ago
  7. rock_mit182 Group Title
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    i mean, they did not give that value

    • 6 months ago
  8. AkashdeepDeb Group Title
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    Yes, you have to find it. Hold on for one second. I'll attach one thing.

    • 6 months ago
  9. rock_mit182 Group Title
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    ok

    • 6 months ago
  10. rock_mit182 Group Title
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    \[|v _{x}|= v _{x0} \cos 40°\] \[|v _{y}|= v _{y0}\sin40°\]

    • 6 months ago
  11. rock_mit182 Group Title
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    \[v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}\]

    • 6 months ago
  12. rock_mit182 Group Title
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    if i use \[\tan 40°\] that would be the initial velocity ?

    • 6 months ago
  13. AkashdeepDeb Group Title
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    Do you have the answer key?

    • 6 months ago
  14. rock_mit182 Group Title
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    im going to check that . . .

    • 6 months ago
  15. rock_mit182 Group Title
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    it seems like it doesn't have the key answer

    • 6 months ago
  16. rock_mit182 Group Title
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    @johnweldon1993

    • 6 months ago
  17. AkashdeepDeb Group Title
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    • 6 months ago
  18. rock_mit182 Group Title
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    @Mashy

    • 6 months ago
  19. AkashdeepDeb Group Title
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    I can explain you the whole thing if you want. But this video, might just clear ALL the queries you have on this topic. Hold on. https://www.youtube.com/watch?v=wDyfD0EAKU8 https://www.youtube.com/watch?v=kc2eCgHtWK0 https://www.youtube.com/watch?v=5WxdXmNrQdc https://www.youtube.com/watch?v=DQRsK_AaSK4 These are the links from an MIT course called 8.01 that I had taken last year. The videos are fantastic! :D

    • 6 months ago
  20. rock_mit182 Group Title
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    i know them but i have to solve this quickly

    • 6 months ago
  21. rock_mit182 Group Title
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    thanks anyway

    • 6 months ago
  22. rock_mit182 Group Title
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    please anyone tell me how should aproach this problem

    • 6 months ago
  23. Mashy Group Title
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    so its just me and u huh? :D

    • 6 months ago
  24. rock_mit182 Group Title
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    lol please help! friend

    • 6 months ago
  25. Mashy Group Title
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    so the basket is vertically 1.05 m high, and 10m away.. and the angle is 40 thats all that matters to us.. we can forget about everything else now |dw:1397055972696:dw|

    • 6 months ago
  26. rock_mit182 Group Title
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    i guess in the horizontal component there is not accelaration

    • 6 months ago
  27. Mashy Group Title
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    correct!

    • 6 months ago
  28. AkashdeepDeb Group Title
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    If you have to solve it quickly. Just use the formula then. y = y_0 + x. tan \(\theta\) - \(\frac{g.x^2}{2.v.cos \theta}\) y = height at landing y_0 = initial height x = distance between both heights This is basically the equation of the parabola in projectile motion. http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

    • 6 months ago
  29. rock_mit182 Group Title
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    so in certain way i was right about the tan 40°

    • 6 months ago
  30. rock_mit182 Group Title
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    the thing is i don't know where that comes from, i mean the equation

    • 6 months ago
  31. AkashdeepDeb Group Title
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    Watch the first video. The equation derivation is very simple. It uses the simple equations of motion. But it considers it in the x and y axis both. :)

    • 6 months ago
  32. Mashy Group Title
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    that comes from the derviation.. just cnosider horizontal and vertical components and u can do it that way as well!

    • 6 months ago
  33. AkashdeepDeb Group Title
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    Yes, you were right, not entirely though. tan theta is not the initial velocity but will be used in calculating it. :)

    • 6 months ago
  34. rock_mit182 Group Title
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    is there a way that not need derivation

    • 6 months ago
  35. Mashy Group Title
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    YES YES.. just decompose buddy lets do this let inital velocity be u then in horizontal we have intial velocity = u cos(40) distance = 10m so time = ? u can calculate lets call that as t now u know in the vertical intial velocity = usin(40) acceleration = -g (upward positive) displacement = +1.05 time = t get an expression .. and solve for u xD tadaaaaa

    • 6 months ago
  36. AkashdeepDeb Group Title
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    http://www.triton.edu/uploadedFiles/Content/Resources_and_Services/Services/Academic_Success_Center/About_Us/projectiles.pdf You do not need derivation for this, if you are allowed to use the equation the directly. The non-calculus proof is linked above ^ . I can tell you WHY we need the equation. As we have an incomplete parabola, due to the height difference, a @Mashy has drawn above, it is necessary to use the parabola equation because the velocity then, can be calculated at that point. :) This is the equation. This equation is the equation used to get all the other equations of height and range. :)

    • 6 months ago
  37. AkashdeepDeb Group Title
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    And there in the equation, y is actually \(\delta y\) giving the change in height. :D

    • 6 months ago
  38. rock_mit182 Group Title
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    |dw:1397056685213:dw|

    • 6 months ago
  39. rock_mit182 Group Title
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    well im going to try in the first way . ..

    • 6 months ago
  40. Mashy Group Title
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    just do the way i told u.. u ll get it :-/

    • 6 months ago
  41. rock_mit182 Group Title
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    |dw:1397057202246:dw|

    • 6 months ago
  42. rock_mit182 Group Title
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    well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?

    • 6 months ago
  43. rock_mit182 Group Title
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    @Mashy

    • 6 months ago
  44. Mashy Group Title
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    i have a hard time understanding ur 'drawings' :P jsut tell me what answer u get :P

    • 6 months ago
  45. rock_mit182 Group Title
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    lol

    • 6 months ago
  46. rock_mit182 Group Title
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    \[u = \frac{ t }{ -g*\sin 40*3.05 }\]

    • 6 months ago
  47. rock_mit182 Group Title
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    is that the equation for the vertical component ?

    • 6 months ago
  48. rock_mit182 Group Title
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    http://www.twiddla.com/1561200

    • 6 months ago
  49. rock_mit182 Group Title
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    i guess im wrong far so far fromget to the answer

    • 6 months ago
  50. rock_mit182 Group Title
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    \[\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m-10mtan40°)2 \cos40° } = v\]

    • 6 months ago
  51. rock_mit182 Group Title
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    no way i couldn't do it

    • 6 months ago
  52. rock_mit182 Group Title
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    @mashy what did i wrong dude ... xD

    • 6 months ago
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