Posted by **Mel** on Friday, September 26, 2008 at 1:31am.

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.

- physics -
**bobpursley**, Friday, September 26, 2008 at 2:31am
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.

- physics -
**mathilda**, Sunday, October 5, 2008 at 1:36pm
How did you manage to get that 55.5 m? Please explain,

thanks

- physics -
**Mitchell**, Tuesday, September 30, 2014 at 11:26pm
y=y0+V0yt+.5Ayt^2

0=30+10t-.5(9.8)t^2

0=4.9t^(2)+10t+30

Via Quadratic Formula

t=-1.656s and/or 3.6969s

since t can't be negative t must equal 3.6969s

x=X0+V0xt+.5Axt^2

x=0+15(3.6969)+.5(0)t^2

x=15(3.6969)

x=55.5m

- physics -
**Nik**, Wednesday, October 1, 2014 at 2:00am
You did a mis-step in your calculation as its ...

y=y0+V0yt+.5Ayt^2

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*

x=15(3.2725)

x=49.5m

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