Question

Question #7935c

Answer 1

I suppose you want the final velocities in the two cases (after the `35.4` m fall).
I would use first the relation connecting distance and time:
`y_f-y_i=v_it+1/2at^2`
with `a=-9.81` aceleration of gravity downwards.
and then the one relating velocities:
`v_f=v_i+at`

Ball 1
`v_i=0` (you let it drop)
So: `0-35.4=0-1/2(9.81)t^2`
`t=2.68` secs
and:
`v_f=0-9.81*2.68=-26.3 m/s` downwards

Ball 2
`v_i=-12.6` downwards
So: `0-35.4=-12.6t-1/2(9.81)t^2` solving the second degree equation:
`t=1.70` secs
and:
`v_f=-12.6-9.81*1.70=-29.3 m/s` downwards