An airplane flying with a velocity...


by spacecadette
Tags: airplane, flying, velocity
spacecadette
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#1
Jan24-09, 03:49 PM
P: 24
[b]1. The problem statement, all variables and given/known data
An airplane is flying with a velocity of 85.0 m/s at an angle of 19.0 degrees above the horizontal. When the plane is a distance 106 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.


[b]2. Relevant equations
How far from the dog will the suitcase land? You can ignore air resistance.

[b]3. The attempt at a solution
I attempted to use Y-Yinitial = Voy(t) - 1/2(g)(t^2)
I solved for t and i got t = 4.651 s.
That answer was incorrect.

I then tried using Vx = (85m/s)(cos19) = 80.37
Vy = (85m/s)(sin19) = 27.67

I then tried adding them using V = sqrt{(80.1^2) - (27.7^2)} = 75.16 m/s

I then did R = Vox(t) but the answer I got wasn't correct.
Help!
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LowlyPion
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#2
Jan24-09, 03:58 PM
HW Helper
P: 5,346
Welcome to PF.

The problem is that the vertical velocity continues to carry it upward before it goes down.

What if you reduced the problem to throwing a ball straight up at 27 m/s and you are standing on a 105m cliff? Find that time and multiply by the horizontal velocity.
spacecadette
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#3
Jan24-09, 04:11 PM
P: 24
I'm still not sure how to solve this =(

LowlyPion
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#4
Jan24-09, 04:16 PM
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An airplane flying with a velocity...


I attempted to use Y-Yinitial = Voy(t) - 1/2(g)(t^2)
I solved for t and i got t = 4.651 s.
You have the right method.

You got the wrong answer from the quadratic.
spacecadette
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#5
Jan24-09, 04:23 PM
P: 24
Ohh! I didn't realize my mistake. I used 0 for Voy instead of 85m/s which is the initial velocity.
I solved the quadratic using the positive value and I got 1.168s. I then multiplied that by 85m/s to find the range and I got 99.28m. Does that sound correct?
LowlyPion
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#6
Jan24-09, 04:45 PM
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Quote Quote by spacecadette View Post
Ohh! I didn't realize my mistake. I used 0 for Voy instead of 85m/s which is the initial velocity.
I solved the quadratic using the positive value and I got 1.168s. I then multiplied that by 85m/s to find the range and I got 99.28m. Does that sound correct?
No.

Your initial vertical velocity is your Voy

Besides your quadratic should yield t to hit the ground. It can't get to the ground in 1 sec from 100 m.
spacecadette
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#7
Jan24-09, 04:47 PM
P: 24
I'm sorry, I'm confused again.
LowlyPion
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#8
Jan24-09, 04:51 PM
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Quote Quote by spacecadette View Post
I'm sorry, I'm confused again.
Write out the equation for your quadratic. Let's see where you are going wrong.
spacecadette
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#9
Jan24-09, 05:01 PM
P: 24
-85 (+ or -) sqrt{(-85^2) - 4(4.9)(-106) all divided by 2(4.9)

I think that when I added the -85 to the problem I ended up with 1.168

Now when I subtracted the -85 to the problem I got 18.51s. Which seems more realistic.
LowlyPion
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#10
Jan24-09, 05:18 PM
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Quote Quote by spacecadette View Post
-85 (+ or -) sqrt{(-85^2) - 4(4.9)(-106) all divided by 2(4.9)

I think that when I added the -85 to the problem I ended up with 1.168

Now when I subtracted the -85 to the problem I got 18.51s. Which seems more realistic.
But 85 is the velocity of the plane and it is not going straight up.

Your vertical component of velocity is 85*Sin19 = 27.67

That should yield 0 = 105 + 27.67*t -4.9*t2

That quadratic yields 8.246 s

I use this on-line calculator btw:
http://www.math.com/students/calcula.../quadratic.htm
spacecadette
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#11
Jan24-09, 05:20 PM
P: 24
Ohhh I understand now. Thank you so much for your help!
spacecadette
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#12
Jan24-09, 05:23 PM
P: 24
How far from the dog will the suitcase land?

I'm still having issues finding the distance.
LowlyPion
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#13
Jan24-09, 05:30 PM
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P: 5,346
Quote Quote by spacecadette View Post
How far from the dog will the suitcase land?

I'm still having issues finding the distance.
That's the easy part now.

You have time to hit the ground.
You have the horizontal velocity projected along the ground ...

Speed * time = distance.


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