## YumYum247 one year ago A soccerball was kicked at an angle of 28 deg to the ground and it landed 39.8m away. a) what was the velocity of the ball when kicked?

1. YumYum247

@Astrophysics just give me a quick hint and leave the rest to me.....

2. Astrophysics

VECTORS PROJECTILE MOTION YAYAYAYA DIAGRAMS

3. YumYum247

|dw:1439958921010:dw|

4. Astrophysics

|dw:1439959033896:dw| I wanted to make an arc

5. YumYum247

just tell me what's the first thing i need to look for....that's all

6. Astrophysics

|dw:1439959053044:dw| now lets make it nice and clean so we know exactly what's going on, we need to find v0 here which is velocity of the ball when it was kicked.

7. Astrophysics

You need to find your velocities at x and y direction, play around with it and apply kinematic formulas

8. YumYum247

yes but for that i need to find the final velocity :(

9. YumYum247

|dw:1439959268885:dw|

10. Astrophysics

We don't actually need it, I have an idea!!!!

11. Astrophysics

Want to hear it

12. YumYum247

yuss!!!!

13. YumYum247

Vertical: Horizontal: a = -9.8m/sec2 dh = 39.8m Dh = 0m/s Vh = ? Vi = ? t = ? Vf = ? t = ?

14. Astrophysics

Ok I was actually thinking it gave us the maximum height

15. YumYum247

nope that's all we got???

16. Astrophysics

Good lets put everything in x and y direction...ok lets see and thing and have fun!

17. YumYum247

sorry i forgot to add the theta in there... Vertical: Horizontal: a = -9.8m/sec2 dh = 39.8m Dh = 0m/s Vh = ? Vi = ? t = ? Vf = ? t = ? Angle = 28deg

18. Astrophysics

x - direction x = 39.8 m $\large v_{x_0} = v_0 \cos (28)$ y-direction a = -g $\large v_{y_0} = v_0 \sin(28)$ so this is what we're given hahahaaa

19. YumYum247

yes....

20. Astrophysics

Ok ok we can actually use the range formula

21. Astrophysics

and find initial velocity instantly XD

22. Astrophysics

$R = \frac{ v_0^2\sin(2 \theta) }{ g }$

23. YumYum247

hey about that..... what does it mean by Vi here, is it the horizontal velo or the vertical???

24. Astrophysics

Oh no that's the resultant velocity |dw:1439960011427:dw| what we're looking for

25. Astrophysics

That can be split into components as shown to figure it out

26. YumYum247

yah cuz that formula says dh = Vi^2 X Sin2theta/a so that's why i was wondering what does that initial velocity means here....

27. YumYum247

horizontal or vertical...cuz if it's the resultant then shouldn't it be Vf^2????

28. Astrophysics

I don't follow, that's the initial velocity, we are looking for the velocity that it was kicked at not at the velocity it landed at

29. YumYum247

|dw:1439960370291:dw|

30. Astrophysics

Oh I see what you're trying to do, no that's not right, because we're splitting the initial velocity into components

31. Astrophysics

|dw:1439960455619:dw|

32. Astrophysics

Those are the components of the initial velocity, does that make sense?

33. Astrophysics

you can label it as i for initial it's the same thing :P

34. YumYum247

those nots confuse me ......i only understand Vf, Vi and Vh

35. Astrophysics

Ok sorry ill fix it

36. YumYum247

Thanks :")

37. Astrophysics

|dw:1439960585668:dw| same thing

38. YumYum247

oh now i see...... :D

39. Astrophysics

Yeah they are just components of the initial velocity :-)

40. Astrophysics

$R = \frac{ v_i^2\sin(2 \theta) }{ g }$

41. YumYum247

ok so where are we going with that formula, are we solving for the Vi there???

42. Astrophysics

Yup!

43. YumYum247

ok so it'll be like this....

44. YumYum247

|dw:1439960730234:dw|

45. YumYum247

do we take the square root??? O_o

46. Astrophysics

Mhm hold on a sec, lets just do the algebra first to avoid confusion the range here is our x or distance which is 39.8 m. It's always best to do algebra first then plug things in especially when you get into deeper physics, it will benefit you. $x= \frac{ v_i^2 \sin(2 \theta) }{ g }$ now solve for the initial velocity

47. Astrophysics

Don't plug anything in just solve for initial velocity

48. Astrophysics

variable

49. YumYum247

ok so rearrange the variables first....

50. Astrophysics

Yup

51. YumYum247

and then plug i nthe numbers....???

52. Astrophysics

Yeah

53. YumYum247

aight thanks man, i really appreciate your help..... Stay cute and blessed!!!! :")

54. Astrophysics

Np, lets just finish it off :P, so what do you get when you solve for initial velocity when you rearrange don't worry about making mistakes, that's how we learn!

55. YumYum247

hold up.....

56. Astrophysics

57. YumYum247

x = Vi^2 X Sin2theta/a x.a = Vi^2 X Sin2theta x.2/sin2Theta = Vi^2

58. Astrophysics

Perfect now just square root it and we're good!

59. YumYum247

|dw:1439961336297:dw|

60. Astrophysics

I'm assuming that 2 in the numerator was meant to be an a haha

61. YumYum247

ok man thanks alot i really appreciate your effort.... stay blessed!! :)

62. Astrophysics

|dw:1439961409567:dw| whats that

63. YumYum247

x

64. YumYum247

or R

65. YumYum247

39.8m

66. Astrophysics

Hmm...our equation was $x= \frac{ v_i^2 \sin(2 \theta) }{ g } \implies v_i = \sqrt{\frac{ gx }{ \sin(2 \theta) }}$

67. Astrophysics

I think you may have added an extra x by mistake

68. YumYum247

yah!!! thats it....!!!

69. YumYum247

no that's just my hand writing :"D

70. YumYum247

ok thanks and good night, sleep tight and don't let the bugs bite :"D

71. Astrophysics

Take care :)

72. YumYum247

too 2 :)