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Objective questions regarding projectile motion - 2 dimension .

  1. Aug 8, 2012 #1
    Objective questions regarding "projectile motion - 2 dimension".

    1. The problem statement, all variables and given/known data
    There are 2 mini problems :

    1. If t1 and t2 be the times of flight from A to B and "θ" be the angle of inclination of AB to the horizontal , then t12 + 2t1t2sinθ + t22 is

    (A.) independent of θ
    (B.) equal to 0
    (C.) dependent on θ
    (D.) none of the above is correct.

    2. If B has horizontal and vertical coordinates x,y referred to A , and the velocity of projection is √(2gh) , and the angle between the two paths at B is a right angle , then B lies on the ellipse ,

    (A.) x2+2y2 = 2hy
    (B) x2 + y2 = 2hy
    (C) 2x2 + 2y2 = 2hy
    (D) x2+2y2 = 2hx

    2. Relevant equations

    http://en.wikipedia.org/wiki/Trajec...tions_at_the_final_position_of_the_projectile

    3. The attempt at a solution

    1. Answer seems to be C but its not !

    2. I don't understand the question seriously. Language seems confusing to me !! :frown:

    Please help...
    Thanks in advance .. :smile:
     
  2. jcsd
  3. Aug 8, 2012 #2

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    I do not understand the first problem at all. What are t1 and t2? Are they times of flight of two projectiles, fired at different angles or different initial speeds or both?

    ehild
     
  4. Aug 8, 2012 #3

    CWatters

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    Deleted.
     
  5. Aug 8, 2012 #4
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Hey ehild ! Thanks for quick reply. :)

    At first I did not understand the problem as language seemed confusing (is). However what the first question says has been illustrated in the below "attatchment :"
     

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  6. Aug 8, 2012 #5

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    I still do not understand. The flight time from point A to B is a single time, not two. So again, what are those times?


    ehild
     
  7. Aug 8, 2012 #6
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Question says that :

    Suppose a body is projected along a parabolic path from the point defined by (0,0). Imagine the x axis as time axis and y axis as position axis. From that (0,0) point , the body reach the point (t1,A). Obviously to reach (t1,A) , the body takes time t1. Now to reach point (t2,B) from here , the body takes time t2 if the time is defined from (0,0). This means there is a time lag of t2-t1 between points (t1,A) and (t2,B).

    Of course , coordinates (t1,A) and (t2,B) lie on a parabola. See the attatchment in above post again.
     
  8. Aug 10, 2012 #7
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Ok , but really , have I to solve that much !

    Ok , lets do the first problem.

    Since range of two projectile is same , then

    Angle of first projectile be x , then other will have 90-x.

    If we join A and B , we make it a symmetric projectile.

    Now how will I proceed ?
     
  9. Aug 10, 2012 #8
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Also t1 = 2usinx/g
    t2 = 2usin(90-x)/g = 2ucosx/g

    Also see the image in the below link :

    http://postimage.org/image/ptms5vrwh/ [Broken]

    Please give some hints to proceed further. I cannot find any way out.
     
    Last edited by a moderator: May 6, 2017
  10. Aug 10, 2012 #9

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    That projectile is not symmetric. You want to hit a point B (X,Y) from the origin. The question is what has to be the launch angle. There are two angles,Φ1 and Φ2, but Φ1 =90-Φ2 is true only for Y=0.

    The angle depends on the position of point B and on the initial speed v.
    Wikipedia provides the formula

    [tex]\tan(\phi)=\frac{v^2 \pm \sqrt{v^4-g^2 X^2-2v^2 gY}}{gx}[/tex]


    http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

    The time the projectile flies from the origin to B is t=X/(vcos(Φ).
    You need to investigate the expression t12+t22+2t1t2sin(θ) where tan(θ)=Y/X.

    You will find the formula 1/cos2(Φ )=1+tan2(Φ ) useful for the derivation.

    ehild
     
    Last edited: Aug 10, 2012
  11. Aug 10, 2012 #10

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    Your equations for t are valid if the projectile final height and initial height are the same. It is not the case in this problem.

    ehild
     
    Last edited by a moderator: May 6, 2017
  12. Aug 12, 2012 #11

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    I show the solution steps a bit further:
    The problem: you want to hit a point B(X,Y) from point A(0,0) with initial speed v. It is possible with two launch angles, Φ1 and Φ2, the times of flight are t1 and t2 respectively.

    Problem 1) says that the line (A,B) encloses the angle θ with the horizontal (x axis), that is, Y/X=tan(θ). Find if the expression

    [itex]E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)[/itex]

    depends on θ.

    Problem 2) says that the initial speed is v=√(2gh), and the trajectories cross each other at 90° angle at B. Show that B is on an ellipse and find out which it is from the given ones.

    If a projectile is aimed to hit a point (X,Y) the tangent of the possible launch angles are

    [itex]\tan(\phi)=\frac{v^2 \pm \sqrt{v^4-g^2 X^2-2v^2 gY}}{gx}[/itex]

    With the notation [itex]C=v^4-g^2 X^2-2v^2 gY[/itex],
    the tangent of the angles are
    [tex]\tan(\phi_1)=\frac{v^2 + \sqrt{C}}{gx}[/tex]
    and
    [tex]\tan(\phi_2)=\frac{v^2 - \sqrt{C}}{gx}[/tex]

    The time of flight from A to B is is
    [tex]t=\frac{X}{v \cos(\phi)}[/tex].
    With that, the expression E is
    [tex]E=t_1^2+t_2^2+2t_1 t_2\sin(\theta)=\frac{X^2}{v^2 \cos^2(\phi_1)}+\frac{X^2}{v^2 \cos^2(\phi_2)}+2\frac{X}{v \cos(\phi_1)}\frac{X}{v \cos(\phi_2)}\sin(\theta)[/tex]

    Using the relation
    [tex]\frac{1}{cos^2(\phi)}=1+\tan^2(\phi)[/tex]

    [tex]E=\frac{X^2}{v^2}\left(1+\tan^2(\phi_1)+1+\tan^2 ( \phi_2)+2\sin(\theta) \sqrt{\left(1+\tan^2(\phi_1)\right) \left(1+\tan^2(\phi_2)\right)} \right) [/tex]
    Plug in the expressions for tan(Φ1) and tan(Φ2), expand and simplify.

    In problem 2), start with the equation of the trajectory:
    [tex]Y=\tan(\phi) X-\frac{gX^2}{2 v^2 \cos^2(\phi)}[/tex].
    the tangent of the trajectory is :
    [tex]Y'=\tan(\phi) -\frac{gX}{v^2 \cos^2(\phi)}[/tex].
    If the two trajectories are perpendicular at B, the product of their tangents is -1.

    [tex]\left( \tan(\phi_1) -\frac{gX}{v^2 \cos^2(\phi_1)} \right) \left ( \tan(\phi_2) -\frac{gX}{v^2 \cos^2(\phi_2)} \right) =-1[/tex].

    Substitute 1+tan2(Φ) for 1/cos2(Φ), and also the expressions for tan(Φ1) and tan(Φ2), expand and simplify.

    ehild
     
  13. Aug 13, 2012 #12
    Re: Objective questions regarding "projectile motion - 2 dimension".

    But sinθ still remains. Answer says that this expression t_1^2+t_2^2+2t_1 t_2\sin(\theta) does not depend on θ.

    What I want to say is this :
    In the below image , one can easily see that this projectile is not symmetric. But arbitrarily assigning line AB as x axis , one can convert it into a symmetric projectile.

    http://postimage.org/image/ptms5vrwh/ [Broken]

    Thanks !! I'll at once do it !!
    :smile:
     
    Last edited by a moderator: May 6, 2017
  14. Aug 13, 2012 #13

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    Both t1 and t2 contain theta, and sin theta cancels with them. Try.
    If you change the direction of gravity, too, perpendicular to AB. Have you got such a great power? :rofl:

    ehild
     
  15. Aug 14, 2012 #14
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Needs to be deleted. Unintentional double posting.
     
    Last edited: Aug 14, 2012
  16. Aug 14, 2012 #15
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Ah !

    I-got-an-idea----idea-animated-animation-smiley-emoticon-000274-facebook.gif

    So resolving g as gcosθ perpendicular to AB and gsinθ (no use till now) , I get ,

    t1 = 2usin(β-θ)/gcos(θ)
    t2 = 2usin(90-(β-θ))/gcos(θ)
    = 2ucos(β-θ)/gcos(θ)

    If I resolved components of g perpendicular to AB and made this projectile symmetric and range of two projectiles are same , the two angles will be complementary.

    So I got two equations for t1 and t2. Now I tried evaluating
    E=t12+t22+2t1t2sin(θ) from here but θ still remained. :frown:

    Am I going correct. You say to plug tan values in terms of x , y and all ? Theta is not canceling up.

    Problem 2 is over. No need to worry about it now. :smile:
     
  17. Aug 14, 2012 #16

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    There are two angles to reach a point (x,y) from the origin, but the angles are complementary only when the initial and final heights are the same.
    In that case, the projectile flies upward for time vsin(β)/g, and back for the same time. So the time of flight is t=2vsin(β)/g.
    The projectile moves with constant velocity vx=vcos(β) in the horizontal direction, so x=vcos(β)2vsin(β)/g=v2/g sin(2β). If you know that the range of the projectile must be x, the launch angle is obtained from the equation sin(2β)=gx/v2. There are two solutions for 2β: 2β1=arcsin(gx/v2) and 2β2=pi-arcsin(gx/v2). That means β2=pi/2-β1.

    If the final height is not the same as the initial height, the time of flight is not 2vsinβ/g, so the above argument can not be applied, the angles are not complimentary. This does not change if you choose a different system of coordinates.

    In your system of coordinates g has both y and x components. The projectile accelerates in the x direction so x≠vcos(β-θ)2vsin(β-θ)/(gcosθ). The angles are not complimentary.

    Imagine θ=80°. Then you need to launch both projectiles at a greater angle but still less than 90°. So both launch angles are less than 10°. Their sum can not be 90°.

    ehild
     

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  18. Aug 18, 2012 #17
    Re: Objective questions regarding "projectile motion - 2 dimension".

    Ok , I agree. But I can also use the equation of range of projectile on an inclined plane. R = u2sin2(β-θ)/gcosθ - gt12/2
    And t1 = 2usin(β-θ)/gcos(θ)

    Like that , I can also express the range in terms of t2.

    Which method will be easy ?
     
  19. Aug 18, 2012 #18

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    Neither one, I guess. I solved the problem with my method, if you do not like it try to solve the other way.

    ehild
     
  20. Aug 19, 2012 #19
    Re: Objective questions regarding "projectile motion - 2 dimension".

    E = [X/cos(θ+z)]2 + [X/cos(θ+y)]2 + 2X2sinθ/cos(θ+z)cos(θ+y)

    On solving , I get :

    E = (X2/(cos2θcos2z + sin2θsin2z - 2cosθcoszsinθsinz)) + (X2/(cos2θcos2y + sin2θsin2y - 2cosθcosysinθsiny)) + 2X2sinθ/(cos2θcoszcosy - sinθcosθcoszcosy - sinθsinzcosθcosy + sin2θ sinzsiny)

    Am I going correct ?
     
  21. Aug 19, 2012 #20

    ehild

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    Re: Objective questions regarding "projectile motion - 2 dimension".

    I do not know what you are doing. What are y and z??????

    ehild
     
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