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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!
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Let me get this straight. You've already derived that the coordinates as a function of time are as follows:
=\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].](http://latex.codecogs.com/png.latex?\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].](http://latex.codecogs.com/png.latex?\displaystyle kt\cos(b)=(1-e^{-kt})\left[\frac{U k \sin(a)}{g+\cos(b)}\right].)
Is that correct?
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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!
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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:
=\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];](http://latex.codecogs.com/png.latex?\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
=(1-e^{-kt})\left(\frac{Uk\sin(a)}{g}+\cos(b)\right)?)
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Yes, that's right! Thanks again!
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Does the projectile start at the origin, which is also the start of the ramp?
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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.
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Ok. I see how it's done. When the projectile hits the ramp, the following condition will be satisfied:
=\dfrac{y(t)}{x(t)},)
right?
Plug in your expressions for )
and .)
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).](http://latex.codecogs.com/png.latex?\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 )
comes from.
Solve for ,)
and you should be able to get the result.
Can you fill in the gaps here?
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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!
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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?
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Thanks again!! I tried to do it again and I arrived at the result! It was an algebraical mess. Thanks for all again!
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You're very welcome. Have a good one!
