Answer
a) $W=27.05J$
b) $PE_0=35.87J$
c) $PE_0=8.82J$
d) $-27J$
Work Step by Step
a) We have $m=0.6kg$, $g=9.8m/s^2$, $h_0=6.1m$ and $h_f=1.5m$
The amount of work done on the ball by its weight is $$W=mg(h_0-h_f)=27.05J$$
b) When the ball is released, its height $h_0=6.1m$, so $$PE_0=mgh_0=35.87J$$
c) When the ball is caught, its height $h_f=1.5m$, so $$PE_f=mgh_f=8.82J$$
d) We have $\Delta PE=PE_f-PE_0=-27.05J=-W$
