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anonymous

  • one year ago

An airplane with a speed of 95.1 m/s is climbing upward at an angle of 32.8 ° with respect to the horizontal. When the plane's altitude is 886 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

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  1. anonymous
    • one year ago
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    My original answers that were both wrong: a. 1230.436935 b. 57.2

  2. welshfella
    • one year ago
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    hmm i cant get my brain into gear at the moment

  3. anonymous
    • one year ago
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    haha, i appreciate your presence!

  4. welshfella
    • one year ago
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    gotta go anyway im afraid

  5. campbell_st
    • one year ago
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    well did you use the tan ratio to find the horizontal distance..?

  6. campbell_st
    • one year ago
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    |dw:1442955509946:dw|

  7. anonymous
    • one year ago
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    |dw:1442955391278:dw| So my X velocity = 79.938 y velocity = 51.51645

  8. campbell_st
    • one year ago
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    well I would have thought you need to horizontal distance, that's part (a) and I don't know vertors...

  9. campbell_st
    • one year ago
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    in part (a) the speed of the plane has no relevance to the horizontal distance

  10. anonymous
    • one year ago
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    Wouldn't it? Because the speed of the plan would be the initial velocity of the package?

  11. anonymous
    • one year ago
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    \[x = 886/\tan(32.8)\] So the plane traveled 1374.8 Meters

  12. campbell_st
    • one year ago
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    that's my answer for part (a) someone will need to help you with part(b)

  13. anonymous
    • one year ago
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    Thank you! I'm gonna keep working at it haha I can't believe I didn't even think of doing that... I ended up with that same answer using kinematics too so yea.

  14. anonymous
    • one year ago
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    Well that's not QUITE part a because it's still asking for the distance the package is falling but making progress!

  15. anonymous
    • one year ago
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    So for the package I got \[-886 = 51.516T - 4.9T^2\] Trying to find the time it's in the air.

  16. anonymous
    • one year ago
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    For the package T = 19.6945

  17. anonymous
    • one year ago
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    I ended up this time with a. 1574.338409 b. 29.37 Degrees I hope someone can confirm this, one of the situations where I only get so many submissions until I lose points lol.

  18. cwrw238
    • one year ago
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    i agree with 1574

  19. anonymous
    • one year ago
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    Thanks for confirming that! Since that's confirmed I can assume that...|dw:1442957461407:dw| so \[\tan(\theta) = \frac{ 886 }{ 1574 }\]

  20. anonymous
    • one year ago
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    Going for it!

  21. anonymous
    • one year ago
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    Ugh stuck on b. 29.37 Degrees didn't work.

  22. cwrw238
    • one year ago
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    No - I'm trying to remember how to find the angle)...

  23. cwrw238
    • one year ago
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    i think you calculate it from the horizontal and vertical components of the velocities at impact time.

  24. anonymous
    • one year ago
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    Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!

  25. cwrw238
    • one year ago
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    the horizontal component will be 95.1 cos 32.8 and you can calculate the vertical one using one of the equations for constant acceleration

  26. anonymous
    • one year ago
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    Are you talking about the X vector? Wouldn't it be the same as the plane which I calculated at 79.9379.

  27. anonymous
    • one year ago
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    I also calculated the final velocity of the y vector to be -28203 m/s so I drew this. |dw:1442958230961:dw|

  28. cwrw238
    • one year ago
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    the x vector would be 79.94 yes

  29. cwrw238
    • one year ago
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    but we need to calculate the vertical component of velocity at time of inpact we could use v = v0 + gt i guess

  30. anonymous
    • one year ago
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    \[V^2=V_0^2+2ax\] \[V^2 = 51.516^2+2(-9.8)(19.6945)\] V=47.6223 Wow ok I must have done something wrong last time or this time because I got WAY different answers. checking now

  31. cwrw238
    • one year ago
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    no thats not right x is not 19.69 - thats t - the time

  32. cwrw238
    • one year ago
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    V = -51.51 + 9.8*19.69

  33. anonymous
    • one year ago
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    Thank you for catching that I was scratching my head ugh been working on this too long. So I got \[V^2 = 51.516^2+(-9.8)(-886) \] Since it's falling -886 V = 141.49 m/s

  34. cwrw238
    • one year ago
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    that gives a velocity of 141.45 which seems feasible

  35. cwrw238
    • one year ago
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    so the required angle = arctan(141.45 / 79.94) = 60.53 degrees

  36. anonymous
    • one year ago
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    So I got now it hits the ground at a 60.53 Degrees

  37. cwrw238
    • one year ago
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    yes its been so long since i did these but i think that's correct

  38. anonymous
    • one year ago
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    Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus

  39. cwrw238
    • one year ago
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    we both got the same answer but doing it different ways so ..

  40. MrNood
    • one year ago
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    haven't done the actual numbers but it seems like most here are missing th epoint

  41. MrNood
    • one year ago
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    |dw:1442959247558:dw| first resolve vx and vy then work out the time to reach h = -886 then work out x distance travelled in that time

  42. cwrw238
    • one year ago
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    did you check on 1574 - was that right?

  43. anonymous
    • one year ago
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    That part was right! Sorry forgot to mention that. Only part b is wrong so far.

  44. anonymous
    • one year ago
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    V_x = 79.9379 V_y = 51.516 So the package will be the same initial velocities? For the package: T = 19.6945

  45. anonymous
    • one year ago
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    Distance the package travels in that time is the x = 1574.33

  46. cwrw238
    • one year ago
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    hmm don't get - it I''m pretty sure you work out the angle using the velocity vectors..

  47. anonymous
    • one year ago
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    Yea I keep getting it too, final velocity of the package before impact is V_x = 79.9379 V_y = -141.49

  48. cwrw238
    • one year ago
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    at impact the Vx will be 79.94 and Vy - well it seems we did that correctly.

  49. MrNood
    • one year ago
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    s = ut + 0.5 at^2 -881 = 51.5 t - 9.81/2 t t=20

  50. cwrw238
    • one year ago
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    its -886

  51. MrNood
    • one year ago
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    if your velocities are correct then angle (wrt horizontal) = tan^-1 vy/vx

  52. cwrw238
    • one year ago
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    yes

  53. anonymous
    • one year ago
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    Maybe there's a rounding error being introduced that's throwing it off I'll double check that I suppose because we all agree on everything so far hah.

  54. cwrw238
    • one year ago
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    anyway gotta go guys its getting late here in the UK

  55. anonymous
    • one year ago
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    Alright, thanks so much for the help!

  56. MrNood
    • one year ago
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    have you been told a default value to use for g? I used 9.8 time = 20 and vx = 80

  57. anonymous
    • one year ago
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    Yes, I was using -9.8 as I was treating ground as 0.

  58. anonymous
    • one year ago
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    Redid it and still am getting 60.53470338 Degrees. Gotta let the dog out, but my professor says all the time that WileyPlus doesn't grade wrong but hmm.

  59. anonymous
    • one year ago
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    I do appreciate the help.

  60. welshfella
    • one year ago
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    I also re did the problem and got 60.5 degrees

  61. welshfella
    • one year ago
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    I just don't see how that can be wrong...

  62. MrNood
    • one year ago
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    OK - I have checked all the figures to 2dp (using 9.81 for g) Here's my input FWIW t = 19.64 s vx = 79.94 vy0 = 51.51 vy1 = -141.16 angle = 60.55 deg

  63. MrNood
    • one year ago
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    and btw horizontal distance travelled = vx * t = 1570m

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