Recent content by chanvincent

They are in opposite direction, so one must be positive while the other one must be negative, let's say v1 is moving to the positive direction, that's mean v2 must be negative... i.e. m_1v_1 + m_2(-v_2) = 0 m_1v_1 = m_2v_2

Okay... that's mean the initial momentum p = 0... And the final momentum depends on two object moving toward opposite direction, that's mean m_1v_1 = m_2v_2 understand?>

Conservaiton of energy always works... so does Conservation of momentum... Since you already have one equation with two unknown, i.e. 22.5 = 1/2m_1v_1^2 + 1/2m_2v_2^2 why don't you apply the conservation of momentum to add one more equation to the problem and make it become 2eqn 2 unkn...

Yes... I meant in your solution, A (0,\sqrt{3}/2) , B (3\sqrt{3}/2, 0) , and P (\sqrt{3}, \sqrt{0.5}) is not collinear..

You know \vec{v} and F_{drag} is not a constant during the flight, do you? There are two major conceptual errors I notice in your solution, First, you separated the x, y, and z component. The drag force ( F_{drag} ) is not a linear combination to \vec{v}_x, \vec{v}_y, \mbox{ and }...

You have made a major algebraic error in finding to coordination of A and B. I notice A, B and P is not colinear

I told you the integral is not easy to do (in your level) dt = \frac{Mdv}{F_{trust}-\frac{1}{2}\rho v^2 C_p A} \int_0^Tdt=\int_0^v \frac{Mdv}{F_{trust}-\frac{1}{2}\rho v^2 C_p A } Any idea how do solve this? Hint: \frac{d}{dx} tanh^{-1}x = \frac{1}{1 - x^2} :-p

Everything is correct up to this point: and your problem is you do not know how to find the x component of the vector... You must know the magnitude of the x component in the unit[/color] vector is not a constant, although the magnitude of the unit[/color] vector is. the x component of...

Since AB=BA, AB can't be -BA unless AB = 0

Yes... you got it... the answer should look like some hyperbolic tangent... just go ahead...

Yes... It is not as easy as you think... I doubt that you could really solve it... but, yes, it could be solved...

m = mass, a = accelration F=ma, => Total Force equals to mass times accelration... understand?

a is accelaration... dv/dt is also accelaration... You havn't learned calculus yet, have you? otherwise you should be familiar with dv/dt... well... Since you claimed that you fully understand the drag equation, I thought you knew calculus... although you do not know how do derive the...

Well, as I said be4, the answer is not simple if the force is not constant. But now you are talking about the constant force.. then the solution IS very simple, You have the drag force F_d = -\frac{1}{2}\rho v^2 C_p A and the force from the trust F_{thrust} = \mbox{constant} , now the...

xez was right... Unless you made a simple assumtion, i.e. the force provided by the vihicle is constant, the answer is anything but far from simple... :smile: