Creative Commons License 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.

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