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Classical Mechanics

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Skydiver with Parachute on a Windy Day

CLMECH-VXGK8O

A skydiver jumps from an airplane with an initial velocity vector of:

$$\vec{v} = (100 \space m/s)\space \hat{x} + 0\space \hat{y} + 0\space \hat{z}$$

Here we have defined the coordinate system with the positive $\hat{y}$ direction pointing north, the positive $\hat{x}$ direction pointing east and the positive $\hat{z}$ direction pointing vertically upwards.

For the first 20.0 seconds after jumping, the skydiver's acceleration vector is:

$$\vec{a} = (-2.00\space m/s^2)\space \hat{x} + (-8.00\space m/s^2)\space \hat{z}$$

For the next 5.00 seconds after opening her parachute, the skydiver has an acceleration vector of:

$$\vec{a} = (-11.0\space m/s^2)\space \hat{x} + (1.00\space m/s^2)\space \hat{y} + (31.0\space m/s^2)\space \hat{z}$$

For the remainder of her fall the skydiver moves with constant velocity, reaching the ground after an additional 60.0 seconds.

How far did the skydiver fall, and what was her horizontal displacement vector?

A

She fell 2310 m; her horizontal displacement was 2080 m at $8.64^\circ$ north of east.

B

She fell 2310 m; her horizontal displacement was 2370 m at $8.64^\circ$ north of east.

C

She fell 2010 m; her horizontal displacement was 1760 m at $0.40^\circ$ north of east.

D

She fell 3530 m; her horizontal displacement was 1560 m at $12.0^\circ$ north of east.