Answer
(a) The first ball.
(b) First ball: $\Delta v=25.3~m/s$
Second ball: $\Delta v=16.5~m/s$
Work Step by Step
(a) The speed of second ball is greater than the speed of the first ball during the fall. But, the increase of speed depends of the time only: $\Delta v=at=gt$. With a lower speed, the first ball takes more time to reach the ground. So, the increase in speed of the first ball is greater than the increase of speed of the second ball.
(b) Let the origin be at the edge of a cliff with the positive direction downward.
First ball: $v_0=0$, $a=g=9.81~m/s^2$, $\Delta x=32.5~m$
$v^2=v_0^2+2a\Delta x$
$v^2=0^2+2(9.81~m/s^2)(32.5~m)$
$v=\sqrt {2(9.81~m/s^2)(32.5~m)}=25.3~m/s$
$\Delta v=v-v_0=25.3~m/s-0=25.3~m/s$
Second ball: $v_0=11.0~m/s$, $a=g=9.81~m/s^2$, $\Delta x=32.5~m$
$v^2=(11.0~m/s)^2+2a\Delta x$
$v^2=(11.0~m/s)^2+2(9.81~m/s^2)(32.5~m)$
$v=\sqrt {(11.0~m/s)^2+2(9.81~m/s^2)(32.5~m)}=27.5~m/s$
$\Delta v=v-v_0=27.5~m/s-11.0~m/s=16.5~m/s$
