A parachutist descending at a speed of 9.40 m/s loses a shoe at an altitude of 31.60 m.?

Question

A parachutist descending at a speed of 9.40 m/s loses a shoe at an altitude of 31.60 m.

(a) When does the shoe reach the ground?

(b) What is the velocity of the shoe just before it hits the ground?

Answer 1

Use the free fall equation:
1/2 g*t^2 + v*t + d0 = d

use 31.6 m for d0
use 0 for d
use -9.807 m/s^2 for g
and use -9.4 m/s for v (remember that the shoe is initially moving at the same velocity as the parachutist)

Solve for t using the quadratic equation.

I got two answers:
1.74 seconds and -3.62 seconds.
The answer you want is the 1.74 seconds. The negative answer is the answer assuming 3.62 seconds ago the shoe was thrown up from the ground at some velocity and on its way down, at 31.6 m above the ground, its velocity was at -9.4 meters per second.

The final velocity is :
v = v0 + g*t

v0 = -9.4
9 = -9.807
t = 1.74 seconds

v is therefore -26.5 m/s

Note, you might need to check these answers. I simplified g to be 10 m/s.

Answer 2

S = ut + 1/2 gt^2 = 31.6 m and u = 9.40 m/sec. Solve for t. By inspection 1.0 < t < 2.0 sec, but you can use the soln for a quadratic equation to be more precise.

v = u + gt; where v is the velocity just before hitting the ground after falling t time and starting with u = 9.4 m/sec initial velocity. If t ~ 1.5 sec, then v ~ 25 m/sec.

You can do the precise math if you need to. The important lesson here is to see how the two SUVAT equations combine to give you the answers for (a) and (b).