Two balls (Ball 1 and Ball 2) are released from the top of a tower. Ball 2 is thrown 3.14 seconds after Ball 1 is dropped. Ball 2 is thrown downward with a velocity of 3.49 m/s. Determine how far Ball 1 has fallen (to two decimal points) by the time Ball 2 is thrown.

Answer 1

If ball1 has been dropping for 3.4 seconds, then ball1 has dropped:

y= vo*t+(1/2)*g t^2 where vo= 0.0 since is was "dropped", not thrown.

or y=.5*9.8*3.49^2= 59.68meters

Answer 2

whoops, I noticed I used 3.4 seconds not 3.14

y= .5*9.8* 3.14^2= 48.31