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.

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

Therefore:
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:

x=x0+v0t+1/2at^2
x=15(3.696)
=55.5 m

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

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.

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

thanks

To graph the motion of the stuntwoman, we can make use of the given information about her initial position, height, and the components of her velocity.

1. x-t graph (horizontal distance vs. time):
Since the helicopter is moving horizontally towards the south, the horizontal distance covered by the stuntwoman will be the same as the horizontal distance covered by the helicopter. She is dropped from the helicopter, so her horizontal position will remain the same as the helicopter's position throughout her fall. Therefore, the x-t graph will be a straight horizontal line.

2. y-t graph (vertical distance vs. time):
The stuntwoman is dropped from the helicopter with an initial position of 30.0 m above the ground. Since she is falling vertically, the y-t graph will be a downward sloping line representing her decreasing height over time. The rate of change of her vertical position with respect to time will be determined by her initial velocity in the upward direction and the acceleration due to gravity pulling her downward.

3. v little x - t graph (velocity in x-direction vs. time):
The horizontal component of the velocity of the stuntwoman remains constant at 15.0 m/s throughout her fall. This means the v little x - t graph will be a straight horizontal line at a value of +15.0 m/s.

4. v little y - t graph (velocity in y-direction vs. time):
The vertical component of the velocity of the stuntwoman changes due to the acceleration caused by gravity. At the moment of being dropped, her initial velocity in the y-direction is 10.0 m/s upward, but as she falls, this velocity decreases at a rate of 9.8 m/s^2 due to gravity. Therefore, the v little y - t graph will be a line sloping downward at a constant rate, representing the decreasing velocity in the y-direction.

To accurately graph these motions, you would need to choose suitable scales for the axes and label them appropriately. Use units of distance (m) for the vertical axes, units of time (s) for the horizontal axes, and indicate the direction and magnitude of the velocity for the v-t graphs.

Please note that the values and scales mentioned here are just for explanation purposes and may not reflect the precise values of the graph.

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