Question

A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What is her average velocity (both magnitude and direction) for the entire fall?
Magnitude m/s
So for the free fall v = (a)(t) and because a = g in this instance I did v = (9.8)(15.0) = 147 m/s
So for the free fall velocity = 147
The second part is confusing, The parachute is open so a != 9.8 because of the wind drag on the chute, does a = 1/2gt^2? Cause if it does I think I can figure it out.

Answer 1

Answer:

$v_{avg} = 6.41 m/s$

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

$d = 625 m + 362 m$

total time is given as

$t = 15 s + 139 s$

so average velocity is given as

$v_{avg} = \frac{625 + 362}{15 + 139}$

$v_{avg} = 6.41 m/s$