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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My original answers that were both wrong: a. 1230.436935 b. 57.2

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0hmm i cant get my brain into gear at the moment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha, i appreciate your presence!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0gotta go anyway im afraid

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0well did you use the tan ratio to find the horizontal distance..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442955509946:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442955391278:dw So my X velocity = 79.938 y velocity = 51.51645

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0well I would have thought you need to horizontal distance, that's part (a) and I don't know vertors...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0in part (a) the speed of the plane has no relevance to the horizontal distance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't it? Because the speed of the plan would be the initial velocity of the package?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x = 886/\tan(32.8)\] So the plane traveled 1374.8 Meters

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0that's my answer for part (a) someone will need to help you with part(b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well that's not QUITE part a because it's still asking for the distance the package is falling but making progress!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for the package I got \[886 = 51.516T  4.9T^2\] Trying to find the time it's in the air.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the package T = 19.6945

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for confirming that! Since that's confirmed I can assume that...dw:1442957461407:dw so \[\tan(\theta) = \frac{ 886 }{ 1574 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ugh stuck on b. 29.37 Degrees didn't work.

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1No  I'm trying to remember how to find the angle)...

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1i think you calculate it from the horizontal and vertical components of the velocities at impact time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1the horizontal component will be 95.1 cos 32.8 and you can calculate the vertical one using one of the equations for constant acceleration

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you talking about the X vector? Wouldn't it be the same as the plane which I calculated at 79.9379.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I also calculated the final velocity of the y vector to be 28203 m/s so I drew this. dw:1442958230961:dw

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1the x vector would be 79.94 yes

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1but we need to calculate the vertical component of velocity at time of inpact we could use v = v0 + gt i guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[V^2=V_0^2+2ax\] \[V^2 = 51.516^2+2(9.8)(19.6945)\] V=47.6223 Wow ok I must have done something wrong last time or this time because I got WAY different answers. checking now

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1no thats not right x is not 19.69  thats t  the time

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1V = 51.51 + 9.8*19.69

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1that gives a velocity of 141.45 which seems feasible

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1so the required angle = arctan(141.45 / 79.94) = 60.53 degrees

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I got now it hits the ground at a 60.53 Degrees

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1yes its been so long since i did these but i think that's correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1we both got the same answer but doing it different ways so ..

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0haven't done the actual numbers but it seems like most here are missing th epoint

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0dw: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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1did you check on 1574  was that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That part was right! Sorry forgot to mention that. Only part b is wrong so far.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0V_x = 79.9379 V_y = 51.516 So the package will be the same initial velocities? For the package: T = 19.6945

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Distance the package travels in that time is the x = 1574.33

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1hmm don't get  it I''m pretty sure you work out the angle using the velocity vectors..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I keep getting it too, final velocity of the package before impact is V_x = 79.9379 V_y = 141.49

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1at impact the Vx will be 79.94 and Vy  well it seems we did that correctly.

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0s = ut + 0.5 at^2 881 = 51.5 t  9.81/2 t t=20

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0if your velocities are correct then angle (wrt horizontal) = tan^1 vy/vx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe 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.

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1anyway gotta go guys its getting late here in the UK

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thanks so much for the help!

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0have you been told a default value to use for g? I used 9.8 time = 20 and vx = 80

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I was using 9.8 as I was treating ground as 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Redid 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do appreciate the help.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0I also re did the problem and got 60.5 degrees

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0I just don't see how that can be wrong...

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0OK  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

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0and btw horizontal distance travelled = vx * t = 1570m
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