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Kinematic Equations Question

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    How long will an arrow be in flight if it is shot at an angle of 25 degrees above the horizontal and hits a target 50.0m away at the same elevation?

    Known: displacement in x-axis: 50.0m
    acceleration in x-axis: 0m/s2

    displacement in y-axis: 0m
    acceleration in y-axis: -9.81m/s2

    theta = 25 degrees

    2. Relevant equations
    This question is at the end of a chapter that deals with kinematic equations, so I'm quite certain that one of the kinematic equations must be employed to solve the problem. Those are:

    vf = vi + a∆t
    ∆d = (vf + vi)/2 * ∆t
    ∆d = vi∆t + 1/2a(∆t)2
    ∆d = vf∆t - 1/2a(∆t)2
    vf2=vi2 + 2a∆d

    3. The attempt at a solution
    Found the y-component of the displacement vector by doing 50m(tan25) = dy, getting a value of 23.315m. I then used pythagorean theorem to get the length of the diagonal displacement, 55.169m. (this is assuming there is no acceleration due to gravity, which is obviously not the case). That was a dead end so I then basically tried using some of the kinematic equations, just as a guess+ check to see if they'd work. One example attempt is:

    ∆dy = (viy)∆t + 1/2 * ay(∆t)2

    The y-displacement is 0m, so I moved the first term on the right side of the equation over to the left side to get

    -(viy)∆t = 1/2 * ay(∆t)2

    Divide each side by t to get

    -(vi) = 1/2 * a∆t

    but I only know one of the three variables and cannot solve. There are a few other, similar attempts to solve the question that ended up not working.

    Some help on this problem would be great, I really appreciate everyone taking the time to read and (hopefully!) solve this =)
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2
    Hi Strider, welcome to PF.

    What do you mean by y-component of the vertical displacement vector? Do you mean y-component of the initoal velocity of the arrow? Remember the y-displacement overall is 0.

    You need to consider the horizontal and vertical components of the motion separately. If the arrow was fired with a speed of v, what would be the horizontal and vertical components of its velocity initially?
     
  4. Dec 8, 2013 #3
    Boylanator- Thanks for the welcome!

    "What do you mean by y-component of the vertical displacement vector? Do you mean y-component of the initoal velocity of the arrow?"

    Sorry for the ambiguity. I meant that I assumed the arrow did not experience any acceleration due to gravity, travelling in a line rather than following a parabolic trajectory. Under that assumption I figured out how high the arrow would have climbed when the range was equal to 50m, given an initial launch angle of 25 degrees. To figure that out I drew a right-angle triangle where one side (which represented the range) was equal to 50m, and the angle between the 50m side and the hypotenuse is 25 degrees. The vertical height that the arrow would have reached was represented by the line opposite the 25 degree angle, and is given by 50m(tan25).

    That is not the y component of the initial velocity of the arrow, as I currently do not know what its initial velocity was.

    "You need to consider the horizontal and vertical components of the motion separately. If the arrow was fired with a speed of v, what would be the horizontal and vertical components of its velocity initially? "

    vx = vcos25
    vy = vsin25
     
  5. Dec 8, 2013 #4
    All objects with mass experience gravity, it's why we can't fire arrows or throw balls into space (not without engines anyway). Remember as well that it is hitting a target on the ground. It IS following a parabolic arc.

    The components are correct. Now that you understand what is going on, can you make any progress?
     
  6. Dec 8, 2013 #5
    I don't think I can make any progress because I don't know what the arrow's initial velocity was. If I had 3 other values that comprise the kinematic equations (final velocity, acceleration, time, or displacement) I would be able to find the arrow's initial velocity and then solve for those components. But I am unable to do that because in either axis I only know two variables.

    In the x-axis I know that displacement is equal to 50m and that acceleration is 0m/s2.

    In the y-axis I know that the displacement equals 0m and acceleration is -9.81m/s2

    It seems that I'm missing some prerequisite information - namely, a third variable included in the kinematic equations - which is needed to solve the problem.
     
  7. Dec 8, 2013 #6
    It can be solved. We assumed an initial speed of v, and you found the components. Can you come up with a simple expression for the time taken to travel 50m in the horizontal?
     
  8. Dec 8, 2013 #7
    speed = distance / time
    time = distance / speed

    Assume an initial speed of v
    vx = vcos25

    time = 50m / vcos25

    Now I just need to find v, so I can in turn get the answer to 50m / vcos25. Presumably v can be found by using one of the 5 KE's?

    To summarize my thought process at the moment: I'm looking for v (initial velocity), I know acceleration is -9.81m/s2 and displacement is 50m. I know time as well but only when expressed in terms of v, so I can't use the time value in order to find the initial velocity.
     
  9. Dec 8, 2013 #8
    Yes your thought process is good. As you noticed earlier there is not enough information to calculate the time or velocity from the horizontal component alone, nor the vertical component alone. You will need to find a link that combines the two, can you think of one?
     
  10. Dec 8, 2013 #9
    A link that combines the horizontal and vertical components. I'm just typing out my thoughts so my apologies for any irrelevant info that may come up:

    I know that v = √[(vcos25)2 + (vsin25)2). While that does take into account both the horizontal and vertical components, it doesn't seem very helpful because I don't have any concrete numbers to put into that equation.

    I already know the acceleration and displacement, so I must find one of the three remaining variables: initial velocity, time, or final velocity.

    Since the arrow starts and ends at the same height, a position-time graph of the arrow would have the same tangent slope magnitude at the beginning (when the arrow is fired) and the end (just before the arrow hits the target). This means that the initial and final velocity are equal in magnitude but opposite in direction. Keeping that in mind I can say that vi = vf, and just change the signs on them as needed once I figure out those values.

    I now have:

    a = -9.81m/s2
    ∆d = 50m
    vi = vf = ?
    t = vicos25

    Plugging those values into the kinematic equation

    vf2 = vi2 + 2a∆d

    vf2 = vf2 + 2a∆d

    vf2 - vf2 = 2a∆d

    0 = 2a∆d

    0 = -9.81m/s2 * 50m

    0 = -490.5m2/s2

    That, um, that doesn't seem quite right.
     
    Last edited: Dec 8, 2013
  11. Dec 8, 2013 #10
    Oh wait one second I may be on to something

    In the equations in my above post, it was incorrect to say vf2= vi2. They have the same magnitude, but opposite directions.

    So from the equation:

    vf2 = vi2 + 2a∆d

    vf2 - vi2 = 2a∆d

    * * *

    vf2 - vi2 = vf2 + vf2

    * * *

    2vf2 = 2a∆d

    2vf2 = -490.5m2/s2

    vf2 = -245.25m2/s2

    This is probably not great form, but the above line will give me a negative number under the square root sign once I root both sides. To avoid that I'm going to retroactively state that forwards and downwards will be represented by positive numbers, allowing gravity's acceleration to be written as 9.81m/s2

    I now have:

    vf2 = 245.25m2/s2

    vf = 15.66045976m/s

    So I have displacement, acceleration, and final velocity. I could find time with the equation

    ∆d=vf∆t - 1/2a(∆t)2

    but I don't know how to isolate for t when it's in both terms and neither term is equal to 0. So instead I'll find vi and use that value to find the horizontal component of velocity, which is constant over time. 50m / vcos25 and I should be good to go.
     
    Last edited: Dec 8, 2013
  12. Dec 8, 2013 #11
    Vf isn't the same as Vi. They have different directions. Looking at the vertical velocities Vf = -Vi. Correct this in your working and come up with an expression for the time.
     
  13. Dec 8, 2013 #12
    I tried solving it my way and got the answer wrong. So back to the drawing board.

    "Vf = -Vi. Correct this in your working and come up with an expression for the time. "

    I need a kinematic equation with both velocities and time.

    ∆d = ([vf + vi]/2) * ∆t

    ∆d = ([vf + (-vf)]/2) * ∆t

    ∆d = (vf - vf) * ∆t

    ∆d = 0 * ∆t

    ∆d/0 = ∆t

    That can't be right either.
     
  14. Dec 8, 2013 #13
    This is the equation for motion with no acceleration, we have acceleration. So we need: vf = vi + at right?
    Remember that we're only considering the vertical motion here so use the vertical velocity component you stated earlier.
     
  15. Dec 8, 2013 #14
    vf = vi + at

    vf - vi = at

    vf - (-vf) = at

    2vf = at

    2vf/a = t

    2(15.66045976m/s)/9.81m/s2 = t

    31.32091952m/s * s2/9.81m = t

    3.192754283s = t

    That answer does make sense (right units, feasible time scale) but it disagrees with the answer in my textbook, which is 2.18s. I double-checked to make sure I was reading the answer for the right question, and it's definitely (and disconcertingly!) listed as 2.18s in my textbook.

    Has the textbook made an error? Or is there some mistake lurking in the text above?
     
  16. Dec 8, 2013 #15
    The final velocity you calculated is incorrect because delta d in the vertical direction is 0.

    From this step: 2vf/a = t

    Swap vf for -vi and use the expression for vi in terms of v that you correctly stated at the beginning.
     
  17. Dec 8, 2013 #16
    This elicits the answer that t=3.19s, which matches what I got in my post above, but differs from the textbook's listed answer.
     
  18. Dec 8, 2013 #17
    Are you remembering these?

    vx = vcos25
    vy = vsin25

    So what is t in terms of v? (not vy or vf or vi)

    Don't worry, you're on your way to 2.18s
     
  19. Dec 8, 2013 #18
    vi = 15.66045976m/s

    vx = 15.66045976m/s * cos25

    vx = 14.193197m/s

    "So what is t in terms of v? (not vy or vf or vi)"

    I'm a tad confused here because we said earlier that v is equal to vi. From post #6, "We assumed an initial speed of v, and you found the components."

    So assuming v = vi

    t = 50m / vcos25

    Since vcos25 = 14.193197m/s

    t = 50m / 14.193197m/s
    = 3.5228s

    Edited to add: if by v you mean the horizontal velocity (vx) then t = 50m /v
     
  20. Dec 8, 2013 #19
    The values for velocity you calculated earlier are incorrect. Thus far we have no numerical value for any velocity.
    At the beginning you derived a simple expression for the time taken to travel to the target in the horizontal frame.
    Since then you have been working with the vertical frame only.

    You got this far correctly in the vertical frame:

    2vf/a = t

    Now you should get rid of vf (vertical final velocity) and replace it with the more general v (initial speed of arrow).

    We want to do this because the horizontal time expression has v as an unknown and we want a similar expression for the vertical time. Do you follow?
     
  21. Dec 8, 2013 #20
    The expression for the time taken to travel to the target in the horizontal frame is 50 / vxcos25

    As for 2vf/a = t

    vertical final velocty = vsin25

    So 2vsin25/a = t

    I don't really follow though, I'm not sure why my previously calculated velocities are incorrect. Nor do I understand why the equation 2vf/a = t doesn't suffice. Isn't the time the arrow takes the same in the vertical frame and horizontal frame?

    Also this method for solving the problem seems very convoluted and long. I acknowledge that's because of my tangents, but nonetheless I think it is not feasible to repeat this method during a test/exam when I don't have 2+ hours and somebody else to help me.
     
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