Question

A parachutist's fall to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. After 3 s, she opens her parachute and experiences an air resistance of 689 N. At what speed (in m/s) is the parachutist falling after 10 s?

Answer 1

The parachutist's speed after 10 seconds is 8.4 m/s.

Step-by-step explanation:

The parachutist's fall to Earth is determined by two opposing forces: gravity and air resistance. Initially, the parachutist experiences a gravitational force of 539 N which causes her to accelerate at a rate of 9.8 m/s² for the first 2 seconds. Using the formula v = at, we can calculate that the parachutist's speed after 2 seconds is 19.6 m/s.

After opening her parachute, the parachutist experiences an air resistance of 615 N which opposes the force of gravity. The net force on the parachutist from 2 s to 10 s is 539 N - 615 N, which is equal to 76 N upward. Since the net force is nonzero, the parachutist's acceleration continues at a constant rate.

Using the formula v = vo + at, where vo is the initial velocity, a is the acceleration, and t is the time, we can calculate the parachutist's speed after 10 s to be 19.6 m/s + (-1.4 m/s²)(8 s) = 8.4 m/s.

Answer 2

Answer:15.8 m/s

Step-by-step explanation:

Given

Gravitational Force
`F_g=605 N`

Air resistance
`F_d=689 N`

mass of Parachutist
`=(605)/(9.8)=61.73 kg`

Velocity after
`3 s`

`v=u+at`

`v=0+9.8* 3`

`v=29.4 m/s`

Now Parachutist opens the parachute

Net for on Parachutist

`605-689=-84`

`84 N` drag force

therefore deceleration
`(84)/(61.73)=1.36 m/s^2`

velocity after 10 s is

`v_1=v+at`

`v_1=29.4-1.36* 10`

`v_1=15.8 m/s`