# projectiles

• May 26th 2011, 07:36 AM
Duke
projectiles
A Rocket R is fired from ground with inital speed s to the horizontal at an angle $\displaystyle \alpha$. There is a resistive force F=-mkv where m is the mass and v is the vector (v1,v2) and K is a constant. Show the equations of motion are

$\displaystyle d/dt(v1)+kv1=0$ and $\displaystyle d/dt(v2)+kv2= -g$

My answer. First in the horizontal direction there is no resultant force so I don't get it
• May 26th 2011, 07:50 AM
TheEmptySet
Quote:

Originally Posted by Duke
A Rocket R is fired from ground with inital speed s to the horizontal at an angle $\displaystyle \alpha$. There is a resistive force F=-mkv where m is the mass and v is the vector (v1,v2) and K is a constant. Show the equations of motion are

$\displaystyle d/dt(v1)+kv1=0$ and $\displaystyle d/dt(v2)+kv2= -g$

My answer. First in the horizontal direction there is no resultant force so I don't get it

That is not true. The air resistance is given by

$\displaystyle F=-mkv=-mk(v_1\mathbf{i}+v_2\mathbf{j})$

This force is proportional to the velocity in both of the x and y directions.

So in the y direction we have by newtons 2nd law

$\displaystyle m\frac{dv_2}{dt}=-mg-kmv_2$

and in the x

$\displaystyle m\frac{dv_1}{dt}=-kmv_1$
• May 26th 2011, 07:56 AM
Duke
How do you know the sign of kmv? Doesn't it needs to be negative.

Also I need to state inital conditions for v and r (vectors)
• May 26th 2011, 08:18 AM
TheEmptySet
First I see why you are confused I factored out the minus signs in the force but forgot to put them in the final answer. Yes I had a typo and have edited the above post.

The initial conditions are implied not explicit given.

Assumption 1 Since we are free to setup our own coordinate system lets start the rocket at the origin. That would give the initial conditions.

$\displaystyle x(0)=0 \quad y(0)=0$

Also since we are launching the rocket its initial velocity should also be zero

$\displaystyle v_1(0)=0 \quad v_2(0)=0$
• May 26th 2011, 08:32 AM
Duke
Thanks. If I define up as positive, do the signs change?
• May 26th 2011, 08:39 AM
TheEmptySet
Quote:

Originally Posted by Duke
Thanks. If I define up as positive, do the signs change?

No this set of equations defines up as positive in the y direction and right as positive in the x direction. You see this from the fact that -mg has a negative sign. If down were positive this would give the term mg.