A person leaps from an airplane 6000 feet above the ground and deploys her/his p

Question

A person leaps from an airplane 6000 feet above the ground and deploys her/his parachute after 2 seconds. Assume that the air resistance both before and after deployment of the parachute results in a deceleration proportional to the person's velocity, but with a different constant of proportionality. Before deployment, take the constant to be 0.15, and after deployment use the value 1.4. (The acceleration due to gravity is 32.2ft/sec^2. Note also that the constants of proportionality given are =k/m, not k.) How fast is the person going when the parachute is deployed? How high off the ground is the person when the parachute is deployed? Approximately (to one decimal place) how much longer does it take the person to reach the ground (give your answer in seconds)?

Explanation / Answer

Solution:

This problem should be solved in two phases from t=0 to t=2 (before deployment) and t>2(after).
In 1st phase,
Int velocity v0 = 0.

at any point in this phase acceleration of the person will be (g - 0.15v)

{since air resistence = k*v, deceleration due to air resistance = (k/m)*v}

Now since a = dv/dt,

The downward acceleration, (g-0.15v) = dv/dt => dt = (1/(g - .15v))dv.

integrating with given initial conditions we get v as a function of time in this phase.

Now the initial condition for the 2nd phase is the final condition for 1st phase,

hence v at t=2 can be found out.

2nd phase is very similar to 1st except for the initial condition and differential equation which is

dv/dt = (g - 1.4v) [downwards]

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