Projectile Motion

• Feb 14th 2011, 12:15 PM
Darkprince
Projectile Motion
I have the following problem:
A projectile of mass m is launched from the origin at time t=0, with speed U, at an angle of elevation a+b to the horizontal. The launch site lies on a ramp that makes an angle b to the horizontal (You are given that 0<a<Pi/4 and 0<b<Pi/4)
During its flight, the projectile is subject to two forces, a gravitational force of magnitude mg, vertically downwards and a drag force -m*k*v, where v is the velocity of the projectile and k>0 is a constant.

a)Derive that the coordinates of the projectile are:

x(t)=(Ucos(a+b))/k)*(1-e^(-kt))
y(t)=(-g*t/k)+((g/k^2)+(Usin(a+b)/k))*(1-e^(-kt))

b) Show that the projectile lands on the ramp when the time t satisfies

k*t*cosb=(1-e^(-kt))*((U*k*sina)/g+cosb)

I have done part a by using Newton's second law and by solving two secodn order differentials equations and by applying the initial conditions.

Now I am trying to do part b and I am struggling to understand where did sina came from, since when the projectile land on the ramp our angle equals to b. I tried substituting in the x,y coordinates of the projectile (a+b)=b but I am not guided anywhere. How is the angle a related with the answer since when the projectile lands on the ramp is at angle b from the origin?? (since the ramp is on angle b)
I will appreciate any help and guidance for solving part b!
• Feb 14th 2011, 12:29 PM
Ackbeet
Let me get this straight. You've already derived that the coordinates as a function of time are as follows:

$\displaystyle x(t)=\frac{U \cos(a+b)}{k(1-e^{-kt})},$

$\displaystyle y(t)=-\frac{gt}{k}+(1-e^{-kt})\left[\frac{g}{k^{2}}+\frac{U\sin(a+b)}{k}\right].$

And you're asked to find an equation that the time must satisfy when it hits the ramp, that equation being

$\displaystyle kt\cos(b)=(1-e^{-kt})\left[\frac{U k \sin(a)}{g+\cos(b)}\right].$

Is that correct?
• Feb 14th 2011, 12:35 PM
Darkprince
No. x(t)=( Ucos(a+b)*(1-e^-k*t) ) / k
Also k*t*cos(b)=(1-e^(-k*t)) * ( (U*k*sina) / g) +cos(b) )

Thanks again for the response!
• Feb 14th 2011, 12:44 PM
Ackbeet
Hmm. You're still not being careful enough with your parentheses. In the second equation there, you have more right parentheses than left parentheses. Is it this:

$\displaystyle x(t)=\frac{U\cos(a+b)(1-e^{-kt})}{k},$

$\displaystyle y(t)=-\frac{gt}{k}+(1-e^{-kt})\left[\frac{g}{k^{2}}+\frac{U\sin(a+b)}{k}\right];$ and you're asked to show

$\displaystyle kt\cos(b)=(1-e^{-kt})\left(\frac{Uk\sin(a)}{g}+\cos(b)\right)?$
• Feb 14th 2011, 12:46 PM
Darkprince
Yes, that's right! Thanks again!
• Feb 14th 2011, 12:47 PM
Ackbeet
Does the projectile start at the origin, which is also the start of the ramp?
• Feb 14th 2011, 12:52 PM
Darkprince
Yes the projectile starts from the origin from an angle a+b. The start of the ramp is at the origin and its angle is b. The projectile starts an angle "a" above the ramp which has angle b and I have to show that the equality holds when the projectile land on the ramp.
• Feb 14th 2011, 01:02 PM
Ackbeet
Ok. I see how it's done. When the projectile hits the ramp, the following condition will be satisfied:

$\tan(b)=\dfrac{y(t)}{x(t)},$ right?

Plug in your expressions for $x(t)$ and $y(t).$ Simplify, simplify, simplify. At one point, you're going to have the expression

$\cos(a+b)\sin(b)-\sin(a+b)\cos(b)=-\left[\sin(a+b)\cos(b)-\cos(a+b)\sin(b)\right]=-\sin((a+b)-(b))=-\sin(a).$

This, incidentally, is where your $\sin(a)$ comes from.

Solve for $kt\cos(b),$ and you should be able to get the result.

Can you fill in the gaps here?
• Feb 14th 2011, 01:22 PM
Darkprince
Thanks for the answer. I am trying the past ten minutes to do the algebraic simplifications but I am somewhere stucked! Is it possible to post me the simplifications because something goes worng with my calculations? Thanks again! I will try to do it again!
• Feb 14th 2011, 01:24 PM
Ackbeet
Well, on this forum we don't hand out the answers. Why don't you post what you've got, and I can see where you're getting stuck?
• Feb 14th 2011, 01:32 PM
Darkprince
Thanks again!! I tried to do it again and I arrived at the result! It was an algebraical mess. Thanks for all again!
• Feb 14th 2011, 01:33 PM
Ackbeet
You're very welcome. Have a good one!