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Projectile motion problem #2

  1. Sep 16, 2012 #1
    If a player jumps with an initial speed of Vo = 7 m/s at an angle of 35, what percent of the jump's range does the player spend in the upper half of the jump?

    So Vo = 7m/s that means that Voy = 7sin60 and Vox = 7cos60 right? I understand (i think) that if I can find how long it takes for the player to reach max height and how long it takes for the player to fall from max height back to the ground, then i should be able to find the answer.
     
  2. jcsd
  3. Sep 16, 2012 #2
    Hello.
    I see your attempt to this problem. To deal with an object going to some velocity at an angle, I think of this as a vector problem. (look in picture)

    Then we need to separate the velocity in components, thus
    [itex]V_{xi} = 7\cos(35)[/itex]
    and
    [itex]V_{yi} = 7\sin(35)[/itex]

    This is clearly a 2D kinematic problem. Thus we need to use
    [itex]x_f = x_i +V_x t +\frac{1}{2}a t^2[/itex]
    where, of course, a = 0 since we assume there is no air resistance and there is nothing that makes object accelerate in the x direction once it has launched. This for the components in the x direction and
    [itex]y_f = y_i +V_y t +\frac{1}{2}g t^2[/itex]
    is for the components in the y direction were
    [itex]g = -9.8 m/s^2[/itex]
    is the acceleration due to gravity.


    Solve this equation for t. For the first equation we have:
    [itex]x_f = 7 t \cos(35)[/itex]
    and
    [itex]y_f = 7 t \sin(35)-\frac{1}{2}9.8 t^2[/itex]

    Since we know that the final position will be at
    [itex](x_max,0)[/itex]
    then
    [itex]y_f = 0[/itex]

    Then solving
    [itex]0 = 7 t \sin(35)-\frac{1}{2}9.8 t^2[/itex]
    for the total time in the air, we get
    [itex]t = \frac{14 \sin(35)}{9.8}[/itex]

    Now the total time required for the player to reach the maximum vertical distance then will be
    half the total time fro the player to land after the jump. Then
    [itex]t_{half} = \frac{1}{2}\frac{14 \sin(35)}{9.8} = \frac{7 \sin(35)}{9.8} [/itex]

    Use this time to find the maximum height:
    [itex]x_{max} = 7 t_{half} \cos(35)[/itex]

    This is not the end of the problem but I think you can do the rest.
     

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    Last edited: Sep 16, 2012
  4. Sep 16, 2012 #3

    Simon Bridge

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    The easy way to puzzle through ballistics problems is to draw the v-t graphs for the x and the y components (separately). Draw the graphs, one under the other, so the time axes coincide.

    If the maximum height is h,
    You want to know the distance traveled horizontally while y≥h.
    The vy-t graph will tell you the times t1 and t2 when y≥h, as well as the time T to come back to the ground again. Using these times and the vx-t graph tells you the total range R and the distance spent with y≥h.


    Hint: look up 60-30-90 triangle - the sine and cosines are easy fractions: no need for calculator. sin(60)=√3/2, cos(60)=1/2
     
  5. Sep 16, 2012 #4
    Sorry I am a bit slow but let me see so:

    "Hello.
    I see your attempt to this problem. To deal with an object going to some velocity at an angle, I think of this as a vector problem. (look in picture)

    Then we need to separate the velocity in components, thus
    Vxi=7cos(35)
    and
    Vyi=7sin(35)"

    Now 7cos(35) gives me 7.73 this is the velocity in the direction of x?

    If that's true then obviously:

    7sin(35) = 4.02 is the velocity in the direction of y?

    If that is right then good I maid some kind of progress but now to your equations I get a little confused:

    Xf = xi + Vxt + 1/2at^2 (Is 7.73 xi? If so then what is Vxt?) and,

    Yf = yi + Vyt + 1/2at^2 (Is 4.02 yi? If so then what is Vyt?)...,

    I am assuming that Xf and Yf are equal to a distance rather than a time?
     
  6. Sep 16, 2012 #5

    Simon Bridge

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    @joseamck: this is a homework forum - please do not do OPs homework for them.

    People learn best when they achieve the results for themselves - some people just need a guiding hand. As you see, vysero only got confused trying to read to understand instead of trying to do to understand.

    Thank you.

    @vysero: you learn better by doing ... if you sketch the vector and a line for the ground you should be able to see for yourself which is the horizontal (x) and which is the vertical (y) component of the velocity. It s just trigonometry - you have done this before.

    formally:
    ##\vec{v}(t)=v_x(t)\hat{\imath}+v_y(t)\hat{\jmath}=v\cos(\theta)\hat{\imath}+v\sin(\theta)\hat{ \jmath}##

    Next step is to sketch the graphs.
    How does ##v_y## vary with time?
    How does ##v_x## vary with time?

    What would be the equation of y(t)?
    What times is ##y=\frac{1}{2}y_{max})##
    How far does the player travel horizontally between those two times?
     
  7. Sep 16, 2012 #6
    Ok if i sketch a triangle with one angle and no side lengths then I have no way of finding anything. Unless im suppose to assume that 7 m/s seconds is a side length.
     
  8. Sep 16, 2012 #7

    Simon Bridge

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    You have two velocity-time graphs to sketch.

    For the x component of the velocity,
    what is the initial velocity?
    what is the final velocity?
    what is the acceleration?
    all these things are given to you.
    except time you don't know - so just label it T.

    see what I mean?

    for the y component of velocity you have another different graph to draw.
    what is the initial velocity in the y direction this time?
    what is the final velocity?
    what is the acceleration?
    all these things are given to you!

    except time you don't know - but you do know it has to be the same as for the x-component graph so give it the same label T and put it the same distance along the t axis.

    You know a lot more than you think you do.
    I'm asking a lot of questions here to guide you - please have a go answering them.
     
  9. Sep 16, 2012 #8
    Max. height=u2/2a


    Vertical velocity at half of the jump,

    v2=u2-2au2/4a

    v2=u2-u2/2

    v=√(7.sin35°)2/2=1.4m/s

    Projectile angle at half jump= arcTan(1.4/(7.Cos35°))=13.88°
    Please note that the horizontal velocity are constant and identical for both elevation.

    The range is proportional to Sin2θ and is proportional to time.
     
    Last edited: Sep 16, 2012
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