## 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. 7 months ago 7 months ago

1. rock_mit182

|dw:1397053258940:dw|

2. rock_mit182

@skullpatrol

3. AkashdeepDeb

You need the equation of a parabola for this. Do you know what the parabola equation for projectile motion is? :)

4. rock_mit182

$y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}$

5. rock_mit182

that one ?

6. rock_mit182

but i don't know about the intial velocity

7. rock_mit182

i mean, they did not give that value

8. AkashdeepDeb

Yes, you have to find it. Hold on for one second. I'll attach one thing.

9. rock_mit182

ok

10. rock_mit182

$|v _{x}|= v _{x0} \cos 40°$ $|v _{y}|= v _{y0}\sin40°$

11. rock_mit182

$v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}$

12. rock_mit182

if i use $\tan 40°$ that would be the initial velocity ?

13. AkashdeepDeb

Do you have the answer key?

14. rock_mit182

im going to check that . . .

15. rock_mit182

it seems like it doesn't have the key answer

16. rock_mit182

@johnweldon1993

17. AkashdeepDeb

18. rock_mit182

@Mashy

19. AkashdeepDeb

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

i know them but i have to solve this quickly

21. rock_mit182

thanks anyway

22. rock_mit182

please anyone tell me how should aproach this problem

23. Mashy

so its just me and u huh? :D

24. rock_mit182

25. Mashy

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

i guess in the horizontal component there is not accelaration

27. Mashy

correct!

28. AkashdeepDeb

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

so in certain way i was right about the tan 40°

30. rock_mit182

the thing is i don't know where that comes from, i mean the equation

31. AkashdeepDeb

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

that comes from the derviation.. just cnosider horizontal and vertical components and u can do it that way as well!

33. AkashdeepDeb

Yes, you were right, not entirely though. tan theta is not the initial velocity but will be used in calculating it. :)

34. rock_mit182

is there a way that not need derivation

35. Mashy

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

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

And there in the equation, y is actually $$\delta y$$ giving the change in height. :D

38. rock_mit182

|dw:1397056685213:dw|

39. rock_mit182

well im going to try in the first way . ..

40. Mashy

just do the way i told u.. u ll get it :-/

41. rock_mit182

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42. rock_mit182

well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?

43. rock_mit182

@Mashy

44. Mashy

i have a hard time understanding ur 'drawings' :P jsut tell me what answer u get :P

45. rock_mit182

lol

46. rock_mit182

$u = \frac{ t }{ -g*\sin 40*3.05 }$

47. rock_mit182

is that the equation for the vertical component ?

48. rock_mit182
49. rock_mit182

i guess im wrong far so far fromget to the answer

50. rock_mit182

$\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m-10mtan40°)2 \cos40° } = v$

51. rock_mit182

no way i couldn't do it

52. rock_mit182

@mashy what did i wrong dude ... xD