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.

```