Hey can someone help with this.

A movie stuntwoman drops from a helicopter that is 30.0 m above the ground and moving with a constant velocity whose components are 10.0 m/s upward and 15.0 m/s horizontal and toward the south. You can ignore air resistance.

What is the horizontal distance(relative to the position of the helicopter when she drops) at which the stuntwoman should have placed the foam mats that break her fall?

I got 55.5 m.

However, the part that I need help with is the graphing. I don't understand how you can graph with this little information.

Draw x-t graph of her motion.

Draw y-t graph of her motion .

Draw v little x - t graph of her motion.

Draw v little y -t graph of her motion.

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  1. x-t...constant horizontal velocity vs time leads to a linear line x on the up axis, t on the horizontal.

    y-t... y=1/2 g t^2 so you see 1/2 of a parabola.

    vx-t constant

    vy=t vy=gt a line.

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  2. How did you manage to get that 55.5 m? Please explain,

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  3. y=y0+V0yt+.5Ayt^2

    Via Quadratic Formula
    t=-1.656s and/or 3.6969s
    since t can't be negative t must equal 3.6969s


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  4. You did a mis-step in your calculation as its ...
    0=30+10t[-.5(9.8)t^2] <- in bracket
    so that will make this ...
    0= (-4.9t^(2))+10t+30

    Via Quadratic Formula
    t=(-1.233s) and/or 3.2725s
    since t can't be negative t must equal 3.2725s

    *steps above*

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  5. Since you are claiming acceleration of gravity to be positive (downwards is positive), then you must record upward values to be negative.

    a of woman= 9.8 m/s^2
    upward velocity of woman= -10 m/s
    helicopter height= -30 m
    horizontal/ southern velocity of woman= 15 m/s
    (south is going downwards, you claimed
    down to be positive).

    First find t:

    y= y0+ v0t+ 1/2at^2
    = -30-10t+4.9t^2
    (rearranging) --> 4.9t^2-10t-30
    \-- use the quadratic formula --/
    t= -1.656 & 3.696
    \--time cannot be negative so use 3.696--/

    Second, find x, the distance the stuntwoman lands relative to the helicopter:

    =55.5 m

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