Question

A skydiver is at an altitude of 1340 m. After 9.3 seconds of free fall, he opens his parachute and finds that the air resistance, FDFD, is given by the formula FD=−bvFD=−bv, where bb is a constant and vv is the velocity. If bb = 68.0 kg/s, and the mass of the skydiver is 65.0 kg, first set up differential equations for the velocity and the position, and then find the speed of the skydiver when the parachute opens.

Answer 1

The differential equation for a skydiver's velocity after deploying a parachute is derived from Newton's second law and the air resistance relation, which leads to the equation dv/dt = g - (b/m)v. The skydiver's speed when the parachute opens can be calculated using the kinematic equation v = u + at, considering the time of free fall and acceleration due to gravity.

Step-by-step explanation:

To set up differential equations for velocity and position of a skydiver, we consider Newton's second law of motion, which states that the force applied on a body is equal to the mass of the body multiplied by its acceleration (F = ma). For a skydiver of mass m = 65.0 kg and air resistance FD = -bv where b = 68.0 kg/s, the differential equation for velocity when the parachute is open can be written as ma = mg - bv. This simplifies to dv/dt = g - (b/m)v, which describes how the velocity v changes over time t. The position equation is derived from the integration of the velocity.

To find the speed of the skydiver when the parachute opens, we can use the kinematic equation v = u + at, where u is the initial velocity (0 m/s for a stationary jump), a is the acceleration due to gravity (approximately 9.81 m/s2), and t is the time of free fall (9.3 seconds). Thus, the speed is v = 0 + (9.81 m/s2) * (9.3 s) which gives us the speed of the skydiver at the moment the parachute is deployed.