Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.

##
Physics 211, Quiz 3A

**
Question 1:
**

Stuntman Dan Koko jumped from the top of the
Vegas World Hotel and Casino, a distance 99.4 meters
above the ground, into an airbag.

How fast was he going when he hit the airbag?

Answer: There are two ways to find his final speed.
Quick method: use the kinematic equation
2 2
v - vo = 2 * a * (x - xo)
We know
initial velocity vo = 0
displacement (x - xo) = -99.4 m
acceleration a = -9.8 m/s^2
Solve for final velocity:
2
v = 2 * (-9.8 m/s^2) * (-99.4 m)
v = sqrt ( 1948 m )
= 44 m/s
The other method takes a little longer. First, we figure
out how long it took him to fall to the airbag; then, we
calculate how fast he was going when he reached it.
Time to fall: use equation
2
(x - xo) = vo * t + 0.5 * a * t
2
-99.4 m = 0 + 0.5 * (-9.8 m/s^2) * t
so
t = sqrt ( 20.3 s^2 )
= 4.5 seconds
Speed after falling for 4.5 seconds is
v = vo + a * t
= 0 + (-9.8 m/s^2) * (4.5 s)
= 44 m/s
One finds the same speed either way.

Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.