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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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
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My original answers that were both wrong: a. 1230.436935 b. 57.2
hmm i cant get my brain into gear at the moment
haha, i appreciate your presence!

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gotta go anyway im afraid
well did you use the tan ratio to find the horizontal distance..?
|dw:1442955509946:dw|
|dw:1442955391278:dw| So my X velocity = 79.938 y velocity = 51.51645
well I would have thought you need to horizontal distance, that's part (a) and I don't know vertors...
in part (a) the speed of the plane has no relevance to the horizontal distance
Wouldn't it? Because the speed of the plan would be the initial velocity of the package?
\[x = 886/\tan(32.8)\] So the plane traveled 1374.8 Meters
that's my answer for part (a) someone will need to help you with part(b)
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.
Well that's not QUITE part a because it's still asking for the distance the package is falling but making progress!
So for the package I got \[-886 = 51.516T - 4.9T^2\] Trying to find the time it's in the air.
For the package T = 19.6945
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.
i agree with 1574
Thanks for confirming that! Since that's confirmed I can assume that...|dw:1442957461407:dw| so \[\tan(\theta) = \frac{ 886 }{ 1574 }\]
Going for it!
Ugh stuck on b. 29.37 Degrees didn't work.
No - I'm trying to remember how to find the angle)...
i think you calculate it from the horizontal and vertical components of the velocities at impact time.
Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!
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
Are you talking about the X vector? Wouldn't it be the same as the plane which I calculated at 79.9379.
I also calculated the final velocity of the y vector to be -28203 m/s so I drew this. |dw:1442958230961:dw|
the x vector would be 79.94 yes
but we need to calculate the vertical component of velocity at time of inpact we could use v = v0 + gt i guess
\[V^2=V_0^2+2ax\] \[V^2 = 51.516^2+2(-9.8)(19.6945)\] V=47.6223 Wow ok I must have done something wrong last time or this time because I got WAY different answers. checking now
no thats not right x is not 19.69 - thats t - the time
V = -51.51 + 9.8*19.69
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
that gives a velocity of 141.45 which seems feasible
so the required angle = arctan(141.45 / 79.94) = 60.53 degrees
So I got now it hits the ground at a 60.53 Degrees
yes its been so long since i did these but i think that's correct
Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus
we both got the same answer but doing it different ways so ..
haven't done the actual numbers but it seems like most here are missing th epoint
|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
did you check on 1574 - was that right?
That part was right! Sorry forgot to mention that. Only part b is wrong so far.
V_x = 79.9379 V_y = 51.516 So the package will be the same initial velocities? For the package: T = 19.6945
Distance the package travels in that time is the x = 1574.33
hmm don't get - it I''m pretty sure you work out the angle using the velocity vectors..
Yea I keep getting it too, final velocity of the package before impact is V_x = 79.9379 V_y = -141.49
at impact the Vx will be 79.94 and Vy - well it seems we did that correctly.
s = ut + 0.5 at^2 -881 = 51.5 t - 9.81/2 t t=20
its -886
if your velocities are correct then angle (wrt horizontal) = tan^-1 vy/vx
yes
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.
anyway gotta go guys its getting late here in the UK
Alright, thanks so much for the help!
have you been told a default value to use for g? I used 9.8 time = 20 and vx = 80
Yes, I was using -9.8 as I was treating ground as 0.
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.
I do appreciate the help.
I also re did the problem and got 60.5 degrees
I just don't see how that can be wrong...
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
and btw horizontal distance travelled = vx * t = 1570m

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