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rock_mit182

  • one year 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
    • one year ago
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    |dw:1397053258940:dw|

  2. rock_mit182
    • one year ago
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    @skullpatrol

  3. AkashdeepDeb
    • one year 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
    • one year ago
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    \[y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]

  5. rock_mit182
    • one year ago
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    that one ?

  6. rock_mit182
    • one year ago
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    but i don't know about the intial velocity

  7. rock_mit182
    • one year ago
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    i mean, they did not give that value

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

  9. rock_mit182
    • one year ago
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    ok

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

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

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

  13. AkashdeepDeb
    • one year ago
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    Do you have the answer key?

  14. rock_mit182
    • one year ago
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    im going to check that . . .

  15. rock_mit182
    • one year ago
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    it seems like it doesn't have the key answer

  16. rock_mit182
    • one year ago
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    @johnweldon1993

  17. AkashdeepDeb
    • one year ago
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  18. rock_mit182
    • one year ago
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    @Mashy

  19. AkashdeepDeb
    • one year 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
    • one year ago
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    i know them but i have to solve this quickly

  21. rock_mit182
    • one year ago
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    thanks anyway

  22. rock_mit182
    • one year ago
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    please anyone tell me how should aproach this problem

  23. Mashy
    • one year ago
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    so its just me and u huh? :D

  24. rock_mit182
    • one year ago
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    lol please help! friend

  25. Mashy
    • one year 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
    • one year ago
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    i guess in the horizontal component there is not accelaration

  27. Mashy
    • one year ago
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    correct!

  28. AkashdeepDeb
    • one year 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
    • one year ago
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    so in certain way i was right about the tan 40°

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

  31. AkashdeepDeb
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    is there a way that not need derivation

  35. Mashy
    • one year 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
    • one year 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
    • one year ago
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    And there in the equation, y is actually \(\delta y\) giving the change in height. :D

  38. rock_mit182
    • one year ago
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    |dw:1397056685213:dw|

  39. rock_mit182
    • one year ago
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    well im going to try in the first way . ..

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

  41. rock_mit182
    • one year ago
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    |dw:1397057202246:dw|

  42. rock_mit182
    • one year 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
    • one year ago
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    @Mashy

  44. Mashy
    • one year 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
    • one year ago
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    lol

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

  47. rock_mit182
    • one year ago
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    is that the equation for the vertical component ?

  48. rock_mit182
    • one year ago
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    http://www.twiddla.com/1561200

  49. rock_mit182
    • one year ago
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    i guess im wrong far so far fromget to the answer

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

  51. rock_mit182
    • one year ago
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    no way i couldn't do it

  52. rock_mit182
    • one year ago
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    @mashy what did i wrong dude ... xD

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