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Projectile motion of marksman

  1. Jan 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A projectile is to be launched by a marksman through an opening in a wall which is 40 m
    away over a flat ground. The lowest part of the opening on the wall is 30 m above the ground
    and the opening’s gap is 10m high. The mechanism for launching the
    projectile generates an initial release velocity of 35 m/s. Ignoring air friction, the effect of
    earth’s curvature, the physical size of the projectile and the width of the wall, find five appropriate launch angles so that the projectile passes through the opening.


    2. Relevant equations

    horizontal motion: x(t)=v*cos(theta)*t
    vertical motion: y(t)=(v*sin(theta)*t)-(0.5*g*t^2)
    time to land on ground: t= 2*v*sin(theta)/g

    3. The attempt at a solution
    i tried solving it by randomly taking bunch of theta angles and find horizontal distance and vertical distance. I believe the horizontal distance should be more than 40 and vertical distance should be between 30-40 m for a given theta angle so that the ball can go through opening. I am able to find various various of horizontal distance but i am getting very small values of vertical distance. I do not know what I am doing wrong.
    Example: theta = 40
    t=(2*35*sin(40))/9.8=4.6s
    x(t)=35*cos(40)*4.6=123m
    y(t)=(35*sin(40)*4.6)-(0.5*9.8*4.6^2)=-0.204?m...?this is wrong, I do not know why?
     
    Last edited: Jan 4, 2016
  2. jcsd
  3. Jan 4, 2016 #2
    Always start with a good picture.
     
  4. Jan 4, 2016 #3
    upload_2016-1-4_21-32-41.png
    there it is
     
  5. Jan 4, 2016 #4

    SteamKing

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    This is your problem. You are throwing various formulas at this situation without considering the physics of the problem.

    Have you made any sketches of the different possible trajectories which the projectile could follow and pass through the opening?

    For instance, in your calculation below, where the launch angle is 40 deg., do you realize that the time of flight of 4.6 s represents a symmetrical path where the projectile is fired from the ground and then returns to the same elevation down range? Isn't the lowest part of the opening 30 m above the ground? How would this launch angle guarantee that the projectile was at least 30 m above ground 40 m down range?

    Physics is much more than merely plugging numbers into a random formula and cranking out an answer.

    Understanding of the problem and how to obtain a reasonable solution must come first.
     
  6. Jan 4, 2016 #5
    That is what I am trying to do. I do not know how to find the appropriate landing time at ground above 30 meter.
     
  7. Jan 4, 2016 #6
    Is the time more constrained by the vertical or the horizontal motion?
     
  8. Jan 4, 2016 #7

    Merlin3189

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    I see others have posted while I was writing, so I've deleted my comments on your first attempt.

    Can I just ask whether you've considered energy?
     
  9. Jan 4, 2016 #8
    No, I have not considered it. I do not see why I need that here.
     
  10. Jan 4, 2016 #9
    Sorry I do not understand your question. Can you please elaborate it?
     
  11. Jan 4, 2016 #10

    Merlin3189

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    I thought consideration of KE and PE would give you a minimum vertical component for the launch velocity and thus a minimum launch angle.
    However I find that is insufficient to find a solution, so perhaps you should persist with the equations of motion.
     
  12. Jan 4, 2016 #11
    Yes true but i do not know what I am doing wrong in using equations of motion. I realized that the time I am calculating is wrong but do not know how to fix it.
     
  13. Jan 4, 2016 #12

    SteamKing

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    In the equation for y(t), you have used two different values for t. Why?
     
  14. Jan 4, 2016 #13
    Oh typo, But i edited it. Answer is still the same..which is wrong..
     
  15. Jan 4, 2016 #14

    SteamKing

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    You've got the basic equations listed in Section 2, but you are not utilizing them correctly.

    You know the location and size of the opening through which you want the projectile to pass through, namely, x = 40 m and 30 m ≤ y ≤ 40 m.

    There are two unknowns to deal with: theta and time.

    You can assume a value of theta (hopefully a value which will lead to a reasonable solution), but what about time?

    Remember, for the projectile to pass through a point with a certain (x,y) value, this must occur at the same instant, so the value of t in x(t) and y(t) must be the same.
     
  16. Jan 4, 2016 #15

    haruspex

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    Rather than trying random solutions, let's try to do it deteministically. The first thing to do is recognise that you need to find the extreme angles. Once you have those it's easy. So you have two projectile problems to solve, one for low point of the window and one for the high point.

    Your "time to land on ground" relevant equation isn't going to be relevant since we do not care when or where the projectile lands.
    There are five SUVAT variables and correspondingly five equations. Each omits one of the five variables. So the first thing you need to do is figure out which equations to use based on which variables are interesting. If you do not know the five SUVAT equations, learn them!

    You want to find the angle, and you know the initial speed, so the initial vertical and horizontal speeds are relevant.
    You also know the vertical and horizontal displacements. That does not give quite enough information to solve either the vertical or horizontal movements. But one more unknown is shared between them: the time.
    So you have two equations, each involving two known values (distance and acceleration) and the same two unknowns (angle and time). Solve the pair of simultaneous equations.
     
  17. Jan 4, 2016 #16
    Yes true, I can assume the value of theta but what are the possible ways to find time?
    How is this idea?
    For example: I assume theta to be 40.
    I can find time through:
    x(t)=v*cos(theta)*t
    so, t=x(t)/v*cos(theta)..... I know horizontal distance and velocity and i assumed theta.
    the time should be same for y(t)
    So, once i have time, i can find x(t) and y(t) and see if x = 40m and 30m ≤ y ≤ 40m. If it is true, that means the at the theta angle, the ball will go through opening.
     
  18. Jan 4, 2016 #17
    But both equations have two unknown variables. Cannot find time or theta unless I have one of them
    So what I understand is:
    For example: I assume theta to be 40.
    I can find time through:
    x(t)=v*cos(theta)*t
    so, t=x(t)/v*cos(theta)..... I know horizontal distance and velocity and i assumed theta.
    the time should be same for y(t)
    So, once i have time, i can find x(t) and y(t) and see if x = 40m and 30m ≤ y ≤ 40m. If it is true, that means the at the theta angle, the ball will go through opening.
     
  19. Jan 4, 2016 #18

    SteamKing

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    Write out your equations for x(t) and y(t) and substitute values for the variables you know, like the initial velocity and the angle, or the location at the end of the trajectory..

    Once you get these equations written out, it takes a little algebra to solve for the value(s) of t which make them true simultaneously.
     
  20. Jan 4, 2016 #19
    I donot know the angle, thats the problem. there are two unknown variables, time and angle.
     
  21. Jan 4, 2016 #20

    SteamKing

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    Like I said before, you can assume a reasonable angle, like you did initially when you took θ = 40°. Once you do that, the only unknown left is time. :wink:
     
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