The distance s that the pilot falls after the parachute opens is given in terms of time as s = 160(1/4-1+e^-t/4)

Velocity is the derivative of s with respect to time and given as:

`(ds)/(dt) = 160*(-1/4)e^(-t/4) = -40*e^(-t/4) = -40/(e^(t/4))`

For t tending to infinity: `lim_(t->oo)(-40/e^(t/4)) = 0`

The...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The distance s that the pilot falls after the parachute opens is given in terms of time as s = 160(1/4-1+e^-t/4)

Velocity is the derivative of s with respect to time and given as:

`(ds)/(dt) = 160*(-1/4)e^(-t/4) = -40*e^(-t/4) = -40/(e^(t/4))`

For t tending to infinity: `lim_(t->oo)(-40/e^(t/4)) = 0`

The terminal velocity of the skydiver which is the constant velocity attained when the air resistance balances the force of gravity is equal to 0 m/s.

The velocity when the pilot jumps off is equal to `lim_(t->0)(-40/e^(t/4)) = -40`

If the time at which the velocity is 5% of the initial velocity is t, `-40*e^(-t/4) = -(0.05*40)` .

Solving for t gives:

`e^(-t/4) = 0.05`

=> `-t/4 = ln 0.05`

=> t = 2.996 s.

At approximately 2.996 s after the pilot jumps off, the velocity is 95% of the initial velocity at jump off.