An airplane flying with a velocity

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Homework Help Overview

The problem involves an airplane flying at a velocity of 85.0 m/s at an angle of 19.0 degrees above the horizontal, with a suitcase dropping from a height of 106 m above the ground. The main question is to determine how far from the dog, which is directly below the plane, the suitcase will land, while ignoring air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the time of flight and the horizontal distance traveled by the suitcase. There are attempts to clarify the initial vertical velocity and the setup of the quadratic equation.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and corrections. Some guidance has been offered regarding the correct use of initial velocities and the setup of equations. There is ongoing exploration of the time to hit the ground and the corresponding horizontal distance.

Contextual Notes

There are indications of confusion regarding the initial vertical velocity and the application of the quadratic formula. Participants are also questioning the assumptions made in their calculations, particularly concerning the vertical motion of the suitcase.

spacecadette
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1. Homework Statement
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.


2. Homework Equations
How far from the dog will the suitcase land? You can ignore air resistance.

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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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.
 
I'm still not sure how to solve this =(
 
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.
 
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?
 
spacecadette said:
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.
 
I'm sorry, I'm confused again.
 
spacecadette said:
I'm sorry, I'm confused again.

Write out the equation for your quadratic. Let's see where you are going wrong.
 
-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.
 
  • #10
spacecadette said:
-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/calculators/source/quadratic.htm
 
  • #11
Ohhh I understand now. Thank you so much for your help!
 
  • #12
How far from the dog will the suitcase land?

I'm still having issues finding the distance.
 
  • #13
spacecadette said:
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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