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

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- anonymous

My original answers that were both wrong:
a. 1230.436935
b. 57.2

- welshfella

hmm i cant get my brain into gear at the moment

- anonymous

haha, i appreciate your presence!

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## More answers

- welshfella

gotta go anyway im afraid

- campbell_st

well did you use the tan ratio to find the horizontal distance..?

- campbell_st

|dw:1442955509946:dw|

- anonymous

|dw:1442955391278:dw|
So my X velocity = 79.938
y velocity = 51.51645

- campbell_st

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

- campbell_st

in part (a) the speed of the plane has no relevance to the horizontal distance

- anonymous

Wouldn't it? Because the speed of the plan would be the initial velocity of the package?

- anonymous

\[x = 886/\tan(32.8)\]
So the plane traveled 1374.8 Meters

- campbell_st

that's my answer for part (a)
someone will need to help you with part(b)

- 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

Well that's not QUITE part a because it's still asking for the distance the package is falling but making progress!

- anonymous

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

- anonymous

For the package T = 19.6945

- 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

i agree with 1574

- anonymous

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

- anonymous

Going for it!

- anonymous

Ugh stuck on
b. 29.37 Degrees didn't work.

- cwrw238

No - I'm trying to remember how to find the angle)...

- cwrw238

i think you calculate it from the horizontal and vertical components of the velocities at impact time.

- anonymous

Hmm ok lemme roll with this for a minute... I THINK THAT'S IT!

- 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

Are you talking about the X vector? Wouldn't it be the same as the plane which I calculated at 79.9379.

- anonymous

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

- cwrw238

the x vector would be 79.94 yes

- cwrw238

but we need to calculate the vertical component of velocity at time of inpact
we could use v = v0 + gt i guess

- 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

no thats not right x is not 19.69 - thats t - the time

- cwrw238

V = -51.51 + 9.8*19.69

- 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

that gives a velocity of 141.45 which seems feasible

- cwrw238

so the required angle = arctan(141.45 / 79.94) = 60.53 degrees

- anonymous

So I got now it hits the ground at a 60.53 Degrees

- cwrw238

yes
its been so long since i did these but i think that's correct

- anonymous

Well, I was very confident as well but still counting it wrong... I really dislike this WileyPlus

- cwrw238

we both got the same answer but doing it different ways so ..

- MrNood

haven't done the actual numbers but it seems like most here are missing th epoint

- 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

did you check on 1574 - was that right?

- anonymous

That part was right! Sorry forgot to mention that. Only part b is wrong so far.

- 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

Distance the package travels in that time is the
x = 1574.33

- cwrw238

hmm don't get - it I''m pretty sure you work out the angle using the velocity vectors..

- anonymous

Yea I keep getting it too, final velocity of the package before impact is
V_x = 79.9379
V_y = -141.49

- cwrw238

at impact the Vx will be 79.94 and Vy - well it seems we did that correctly.

- MrNood

s = ut + 0.5 at^2
-881 = 51.5 t - 9.81/2 t
t=20

- cwrw238

its -886

- MrNood

if your velocities are correct then
angle (wrt horizontal) = tan^-1 vy/vx

- cwrw238

yes

- 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

anyway gotta go guys
its getting late here in the UK

- anonymous

Alright, thanks so much for the help!

- MrNood

have you been told a default value to use for g?
I used 9.8
time = 20 and vx = 80

- anonymous

Yes, I was using -9.8 as I was treating ground as 0.

- 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

I do appreciate the help.

- welshfella

I also re did the problem and got 60.5 degrees

- welshfella

I just don't see how that can be wrong...

- 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

and btw
horizontal distance travelled = vx * t = 1570m

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