
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)*(1e^(kt))
y(t)=(g*t/k)+((g/k^2)+(Usin(a+b)/k))*(1e^(kt))
b) Show that the projectile lands on the ramp when the time t satisfies
k*t*cosb=(1e^(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!

Let me get this straight. You've already derived that the coordinates as a function of time are as follows:
$\displaystyle \displaystyle x(t)=\frac{U \cos(a+b)}{k(1e^{kt})},$
$\displaystyle \displaystyle y(t)=\frac{gt}{k}+(1e^{kt})\left[\frac{g}{k^{2}}+\frac{U\sin(a+b)}{k}\right].$
And you're asked to find an equation that the time must satisfy when it hits the ramp, that equation being
$\displaystyle \displaystyle kt\cos(b)=(1e^{kt})\left[\frac{U k \sin(a)}{g+\cos(b)}\right].$
Is that correct?

No. x(t)=( Ucos(a+b)*(1e^k*t) ) / k
Also k*t*cos(b)=(1e^(k*t)) * ( (U*k*sina) / g) +cos(b) )
Thanks again for the response!

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

Yes, that's right! Thanks again!

Does the projectile start at the origin, which is also the start of the ramp?

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.

Ok. I see how it's done. When the projectile hits the ramp, the following condition will be satisfied:
$\displaystyle \tan(b)=\dfrac{y(t)}{x(t)},$ right?
Plug in your expressions for $\displaystyle x(t)$ and $\displaystyle y(t).$ Simplify, simplify, simplify. At one point, you're going to have the expression
$\displaystyle \cos(a+b)\sin(b)\sin(a+b)\cos(b)=\left[\sin(a+b)\cos(b)\cos(a+b)\sin(b)\right]=\sin((a+b)(b))=\sin(a).$
This, incidentally, is where your $\displaystyle \sin(a)$ comes from.
Solve for $\displaystyle kt\cos(b),$ and you should be able to get the result.
Can you fill in the gaps here?

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!

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?

Thanks again!! I tried to do it again and I arrived at the result! It was an algebraical mess. Thanks for all again!

You're very welcome. Have a good one!