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rock_mit182

  • 8 months ago

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.

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  1. rock_mit182
    • 8 months ago
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    |dw:1397053258940:dw|

  2. rock_mit182
    • 8 months ago
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    @skullpatrol

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

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

  5. rock_mit182
    • 8 months ago
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    that one ?

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

  7. rock_mit182
    • 8 months ago
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    i mean, they did not give that value

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

  9. rock_mit182
    • 8 months ago
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    ok

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

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

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

  13. AkashdeepDeb
    • 8 months ago
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    Do you have the answer key?

  14. rock_mit182
    • 8 months ago
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    im going to check that . . .

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

  16. rock_mit182
    • 8 months ago
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    @johnweldon1993

  17. AkashdeepDeb
    • 8 months ago
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  18. rock_mit182
    • 8 months ago
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    @Mashy

  19. AkashdeepDeb
    • 8 months ago
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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

  20. rock_mit182
    • 8 months ago
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    i know them but i have to solve this quickly

  21. rock_mit182
    • 8 months ago
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    thanks anyway

  22. rock_mit182
    • 8 months ago
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    please anyone tell me how should aproach this problem

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

  24. rock_mit182
    • 8 months ago
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    lol please help! friend

  25. Mashy
    • 8 months ago
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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|

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

  27. Mashy
    • 8 months ago
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    correct!

  28. AkashdeepDeb
    • 8 months ago
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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

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

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

  31. AkashdeepDeb
    • 8 months ago
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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. :)

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

  33. AkashdeepDeb
    • 8 months ago
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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. :)

  34. rock_mit182
    • 8 months ago
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    is there a way that not need derivation

  35. Mashy
    • 8 months ago
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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

  36. AkashdeepDeb
    • 8 months ago
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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. :)

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

  38. rock_mit182
    • 8 months ago
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    |dw:1397056685213:dw|

  39. rock_mit182
    • 8 months ago
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    well im going to try in the first way . ..

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

  41. rock_mit182
    • 8 months ago
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    |dw:1397057202246:dw|

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

  43. rock_mit182
    • 8 months ago
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    @Mashy

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

  45. rock_mit182
    • 8 months ago
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    lol

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

  47. rock_mit182
    • 8 months ago
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    is that the equation for the vertical component ?

  48. rock_mit182
    • 8 months ago
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    http://www.twiddla.com/1561200

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

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

  51. rock_mit182
    • 8 months ago
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    no way i couldn't do it

  52. rock_mit182
    • 8 months ago
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    @mashy what did i wrong dude ... xD

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