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  1. #1
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    Reps for physics help

    1) An airplane with a speed of 86.1 m/s is climbing upward at an angle of 31.5 ? with respect to the horizontal. When the plane's altitude is 805 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

    The answer to part (a) is 1313.6m but I can't get the answer to (b).

    2) In the javelin throw at a track-and-field event, the javelin is launched at a speed of 23.0 m/s at an angle of 31.6 ? above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 31.6 ? at launch to 16.0 ??

    Any help with these two would be appreciated.
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  2. #2
    Registered User Tootella's Avatar
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    Originally Posted by ZekeXX View Post
    1) An airplane with a speed of 86.1 m/s is climbing upward at an angle of 31.5 ? with respect to the horizontal. When the plane's altitude is 805 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

    The answer to part (a) is 1313.6m but I can't get the answer to (b).

    2) In the javelin throw at a track-and-field event, the javelin is launched at a speed of 23.0 m/s at an angle of 31.6 ? above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 31.6 ? at launch to 16.0 ??

    Any help with these two would be appreciated.

    for b)

    the x component will be the same as the plane's x component of velocity.

    the y component will have to be found using y= 1/2gt^2 since you know the height, you can use that to find time. then when you have time, you can use it to find the final velocity just before it hits the ground with the equation (v-final)^2= -2g(delta time)

    so when you have the x and y components, then you can use trig to figure out the angle. x will be a positive x value and the y should be a -y value, so the angle should probably be in the range of 270-360 degrees, or 0-90 below the horizontal.

    does that make sense?
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  3. #3
    BEAST MODE nickrut's Avatar
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    Originally Posted by ZekeXX View Post
    1) An airplane with a speed of 86.1 m/s is climbing upward at an angle of 31.5 ? with respect to the horizontal. When the plane's altitude is 805 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

    The answer to part (a) is 1313.6m but I can't get the answer to (b).

    2) In the javelin throw at a track-and-field event, the javelin is launched at a speed of 23.0 m/s at an angle of 31.6 ? above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 31.6 ? at launch to 16.0 ??

    Any help with these two would be appreciated.
    Answer to B is tan-1(805/1313.6).
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  4. #4
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    Originally Posted by ZekeXX View Post
    2) In the javelin throw at a track-and-field event, the javelin is launched at a speed of 23.0 m/s at an angle of 31.6 ? above the horizontal.
    Epicly failed throw. Optimum angle is 45?.



    Originally Posted by Tootella View Post
    for b)

    the x component will be the same as the plane's x component of velocity.

    the y component will have to be found using y= 1/2gt^2 since you know the height, you can use that to find time. then when you have time, you can use it to find the final velocity just before it hits the ground with the equation (v-final)^2= -2g(delta time)

    so when you have the x and y components, then you can use trig to figure out the angle. x will be a positive x value and the y should be a -y value, so the angle should probably be in the range of 270-360 degrees, or 0-90 below the horizontal.

    does that make sense?
    How do you know this...
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  5. #5
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    niether of these answers were correct.
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  6. #6
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  7. #7
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    any serious replies?
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  8. #8
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    i can do number 2 for you, for sure, just one sec

    for number 1b, i cant figure if its a trick question or not, im thinking the package just falls vertically but i have a feeling im way wrong
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  9. #9
    Diamond Member Status gpg_prodigy's Avatar
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    what chapter are u on if u tell me maybe i can help
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  10. #10
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    alright 2 is harder than i expected i think it can be solved using related rates though

    edit: 2 is really easy

    you know

    tan (theta) = y comp of velocity/ x comp of velocity

    since you solved 1a ill assume you can resolve vectors, so resolve the inital situation you will get x comp = 19.58, y comp = 12.05

    but the y comp changes by 9.8m/s^2, x comp stays the same obviously

    so you have

    tan(theta) = (12.05-9.8t)/(19.58)

    plug in 16 for theta and solve for t

    t=0.65s
    Last edited by skinnycalves; 02-02-2009 at 09:28 PM.
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  11. #11
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    1b is merely a projectile motion problem, you better log back on and see the work i did for you ***got, i spend a good 10-20 minutes thinking back to the last time i took physics
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  12. #12
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    when the package is released its horizontal speed is that of the plane, which since you solved 1a, you know that x comp velocity = 73.41 m/s, imagine at that point in time hte package is being thrown from the cliff at 73.41 m/s (neglecting air resistance this should not change at all during the flight path), and its y comp velocity initial = 0 m/s but that is subject to an acceleration by gravity

    distance in y direction=.5gt^2 (since velocity y initial is 0, dont even include it)
    805=4.9t^2
    t= 12.8 seconds

    the ball travels for 12.8 seconds before hitting the ground

    so horizontal d = 73.41(t) = 73.41 (12.8) = 939.648 meters from release point

    add that to 1313 from the takeoff point you get 2252.648 meters
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  13. #13
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    if this was for some internet quiz due at 12 o clock, im sorry for not stumbling upon it earlier
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