That's it, I'm putting down what I originally had as my answer:
"Plane's do not drop packages. If this plane dropped the package, then the package would be destroyed."
But, we could assume the package is dropped from an altitude of 900m?
That's what I had thought as well, that the height of the plane was fixed.
If i go with this approach, then the plane doesn't have two velocity components? Only one because the height is unchanging. Which would mean the package's velocity in the y component is due to gravity; it would be in...
I see what you are saying, in the initial velocity it gt is not involved, but AFTER it is dropped, yes.
I think I have worked out a solution.
Bare with me.
1. Vpack (y component) = Vplane*sinθ + gt
2. ((Vpack y)^2 - (Vpack nought y)^2)/(2g) = dy (altitude from which the package has to...
@lewando, the x-axis would be the distance's between the nose of the plane and the rear of the ship. The y-axis would be the difference in altitudes.
Okay, I drew a diagram
the velocity of the plane is given but it has an x and v component.
Vplane = 180 m/s
Thus the intial velocity...
Homework Statement
A plane flies at 900m and has to deliver a package. It flies at 180 m/s North at 15° below the horizontal. The ships velocity is 40 m/s and it is traveling due north. At what horizontal distance from the ship must the package be dropped for it to land on the ship?Homework...
I have not come across any of the terminology you have just said.
I doubt I would be allowed to use that method.
Could you describe it in another way or comment on my attempt?
Homework Statement
I have to derive an equation with the following variables:
T= period
m= mass
Theta= angle
l= length
and I was told to think of the period as the perimeter of a circle.Homework Equations
The Attempt at a Solution
C=2\pi r,C/2=\pi r,d=\pi r,(at^2)/2=\pi r,t=\sqrt{\frac{2\pi...
In the diagram above, m1 = 285g, m2 = 755g, theta = 10.0 degrees, and mu = .047.
If system is released from rest, how long does it take m2 to reach the end of table.
Assum m1 can fall atleast 2.50 m.
The tension (resultant and tension for the hanging box) was calculated using the weight of the hanging box.
Also, I'm in a Trig based physics course and we CAN'T do derivatives, actually we're not supposed to.
I meant to say that the tension of the falling box is equal to the resultant tension of the sliding box. The tension in the x direction would be cos (10) * Tension of falling box.
Homework Statement
http://img813.imageshack.us/img813/594/75309500.png
http://img813.imageshack.us/img813/7166/51848816.png
I have to find acceleration to calculate how long it takes the heavier box to reach the end of the table.
Homework Equations
The Attempt at a Solution...