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# Force diagrams, vectors and equations of motion

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A parachutist and her parachute have a combined mass of 80 kg. She steps out from an aircraft travelling horizontally at 90m s−1 at a height of 300m above the ground, and falls for 5 seconds before instantaneously opening her parachute. In the free fall phase of the motion, take the origin of coordinates to be a point on the ground vertically below where the parachutist jumped from the aircraft. Model the parachutist as a particle, and ignore air resistance. You may assume that the velocity of the parachutist relative to the
Air-craft is initially zero.

I would like to know what a force diagram for the parachutist would look like with the corresponding unit vectors. Could you please clearly deﬁne a system of coordinates?

What would the initial position vector and velocity of the parachutist be in terms of these unit vectors?

How would I determine the diﬀerential equation of motion of the parachutist and then derive expressions for her velocity v(t) and position r(t).

How would I calculate the speed and the direction of the parachutist's motion with respect to the horizontal at the instant when the parachute opens, and what would be her vertical height at this instant?

If I was to redeﬁne the origin for the phase after the parachute has opened to be at the point on the ground vertically below that at which the parachute opens, and then model the parachutist and parachute as a sphere of eﬀective diameter 3m, what are the position, speed and direction of motion of the parachutist at the instant when the parachute opens in terms of the new coordinates? (Assuming that the quadratic model of air resistance applies in addition to the force due to gravity)

If you make use of any formulas could you clearly state them and why they may be used please

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Force diagrams, vectors and equations of motion are examined.

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Newton's second law states that the force F is proportional to the acceleration a:
(1.1)
where M is the mass of the object.
Since this is a vector equation, it must be true for each dimension separately:
(1.2)
Where
(1.3)
The acceleration is simply the second derivative with respect to time of the position vector r:
(1.4)
Since
(1.5)
We get:
(1.6)
Thus, the equations of motion in two dimensions are:
(1.7)
During the first phase (the free fall) the only force acting on the parachutist is gravity :

There is no ...

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