A skydiver, weighing 210 lb (including equipment) falls vertically downward from

Question

A skydiver, weighing 210 lb (including equipment) falls vertically downward from an altitude of 6000 ft and opens the parachute after 10 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.75|v| when the parachute is closed and is of magnitude 14|v| when the parachute is open, where the velocity v is measured in ft/s

Determine how long the sky diver is in the air after the parachute opens. Round your answers to two decimal places

Explanation / Answer

Here I'm assuming that the reader has knowledge of basic integration techniques, pariicularly definite integrals. I'll be focusing more on the important plot points pertaining to the solution, which are extremely crucial. However, if the reader requires, I'll provide a detailed handwritten solution, complete with all the calculations.

First of all, we choose our directions. As there is only the vertical direction in which we are interested (assuming the skydiver comes down vertically), we will assume that vertically downwards is the positive z direction, with z=0 the point where the skydiver jumps off the plane, with zero velocity (also assumed). Therefore, at t=0, z=0 and v=0.

As v is in the downward direction (+z), |v| = v.

Then, the motion is divided into two parts -

1) Part where the force balance is ma = mg - 0.75v. Therefore, a = g - (0.75/m)v. This is for 10 seconds only.

2) Part where the force balance is ma = mg - 14v. Therefore, a = g - (14/m)v.

a = acceleration of the skydiver in the downward vertical direction.

g = gravitational acceleration, also acting in the downward vertical direction; constant.

m = mass of skydiver.

Now, the solution involves finding the distance traversed in the first 10 seconds (z1) of the motion, and the velocity at the end of those 10 seconds (v1) using a = dv/dt = g - (0.75/m)v, obtaining v as a function of t. Subsequently, the distance traversed in those 10 seconds is found using dz/dt = v. The values obtained are - v1 = 316.06 ft/s, z1 = 1589.72 ft.

In the next step, the velocity for the second part of the motion is found out as a function of time using a = dv/dt = g - (14/m)v, using the lower limit of the integral of dv as v1. Again, subsequently, dz/dt = v or dz = vdt is used to find the time taken by the skydiver to reach the ground after he/she opens the parachute, using the lower and upper limits on the integral of dz as z1 and 6000 ft, respectively. There is a tricky part here in finding the desired time, as the equation for t which comes out involves terms linear and exponential in t. The trick is to assume that the time taken is long enough for the exponential term [exp(-bt); b being a constant = 14/m] to become negligible.

Thus, the final answer comes out to be - t = 14.31 seconds.

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