anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
My original answers that were both wrong: a. 1230.436935 b. 57.2
welshfella
  • welshfella
hmm i cant get my brain into gear at the moment
anonymous
  • anonymous
haha, i appreciate your presence!

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welshfella
  • welshfella
gotta go anyway im afraid
campbell_st
  • campbell_st
well did you use the tan ratio to find the horizontal distance..?
campbell_st
  • campbell_st
|dw:1442955509946:dw|
anonymous
  • anonymous
|dw:1442955391278:dw| So my X velocity = 79.938 y velocity = 51.51645
campbell_st
  • campbell_st
well I would have thought you need to horizontal distance, that's part (a) and I don't know vertors...
campbell_st
  • campbell_st
in part (a) the speed of the plane has no relevance to the horizontal distance
anonymous
  • anonymous
Wouldn't it? Because the speed of the plan would be the initial velocity of the package?
anonymous
  • anonymous
\[x = 886/\tan(32.8)\] So the plane traveled 1374.8 Meters
campbell_st
  • campbell_st
that's my answer for part (a) someone will need to help you with part(b)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Well that's not QUITE part a because it's still asking for the distance the package is falling but making progress!
anonymous
  • anonymous
So for the package I got \[-886 = 51.516T - 4.9T^2\] Trying to find the time it's in the air.
anonymous
  • anonymous
For the package T = 19.6945
anonymous
  • anonymous
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.
cwrw238
  • cwrw238
i agree with 1574
anonymous
  • anonymous
Thanks for confirming that! Since that's confirmed I can assume that...|dw:1442957461407:dw| so \[\tan(\theta) = \frac{ 886 }{ 1574 }\]
anonymous
  • anonymous
Going for it!
anonymous
  • anonymous
Ugh stuck on b. 29.37 Degrees didn't work.
cwrw238
  • cwrw238
No - I'm trying to remember how to find the angle)...
cwrw238
  • cwrw238
i think you calculate it from the horizontal and vertical components of the velocities at impact time.
anonymous
  • anonymous
Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!
cwrw238
  • cwrw238
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
anonymous
  • anonymous
Are you talking about the X vector? Wouldn't it be the same as the plane which I calculated at 79.9379.
anonymous
  • anonymous
I also calculated the final velocity of the y vector to be -28203 m/s so I drew this. |dw:1442958230961:dw|
cwrw238
  • cwrw238
the x vector would be 79.94 yes
cwrw238
  • cwrw238
but we need to calculate the vertical component of velocity at time of inpact we could use v = v0 + gt i guess
anonymous
  • anonymous
\[V^2=V_0^2+2ax\] \[V^2 = 51.516^2+2(-9.8)(19.6945)\] 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
  • cwrw238
no thats not right x is not 19.69 - thats t - the time
cwrw238
  • cwrw238
V = -51.51 + 9.8*19.69
anonymous
  • anonymous
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
cwrw238
  • cwrw238
that gives a velocity of 141.45 which seems feasible
cwrw238
  • cwrw238
so the required angle = arctan(141.45 / 79.94) = 60.53 degrees
anonymous
  • anonymous
So I got now it hits the ground at a 60.53 Degrees
cwrw238
  • cwrw238
yes its been so long since i did these but i think that's correct
anonymous
  • anonymous
Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus
cwrw238
  • cwrw238
we both got the same answer but doing it different ways so ..
MrNood
  • MrNood
haven't done the actual numbers but it seems like most here are missing th epoint
MrNood
  • MrNood
|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
cwrw238
  • cwrw238
did you check on 1574 - was that right?
anonymous
  • anonymous
That part was right! Sorry forgot to mention that. Only part b is wrong so far.
anonymous
  • anonymous
V_x = 79.9379 V_y = 51.516 So the package will be the same initial velocities? For the package: T = 19.6945
anonymous
  • anonymous
Distance the package travels in that time is the x = 1574.33
cwrw238
  • cwrw238
hmm don't get - it I''m pretty sure you work out the angle using the velocity vectors..
anonymous
  • anonymous
Yea I keep getting it too, final velocity of the package before impact is V_x = 79.9379 V_y = -141.49
cwrw238
  • cwrw238
at impact the Vx will be 79.94 and Vy - well it seems we did that correctly.
MrNood
  • MrNood
s = ut + 0.5 at^2 -881 = 51.5 t - 9.81/2 t t=20
cwrw238
  • cwrw238
its -886
MrNood
  • MrNood
if your velocities are correct then angle (wrt horizontal) = tan^-1 vy/vx
cwrw238
  • cwrw238
yes
anonymous
  • anonymous
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.
cwrw238
  • cwrw238
anyway gotta go guys its getting late here in the UK
anonymous
  • anonymous
Alright, thanks so much for the help!
MrNood
  • MrNood
have you been told a default value to use for g? I used 9.8 time = 20 and vx = 80
anonymous
  • anonymous
Yes, I was using -9.8 as I was treating ground as 0.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I do appreciate the help.
welshfella
  • welshfella
I also re did the problem and got 60.5 degrees
welshfella
  • welshfella
I just don't see how that can be wrong...
MrNood
  • MrNood
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
MrNood
  • MrNood
and btw horizontal distance travelled = vx * t = 1570m

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