Answer
$\Delta U=mg\left( v_{0y}t-g\dfrac {t^{2}}{2}\right) \approx -32.4J$
Work Step by Step
Assuming that ball is thrown from initial height $H_{0}$ with a vertical component of velocity $v_{0y} $ at $t=0$ then at any giving time during free fall the height of the ball will be: $H=H_{0}+v_{0y}t-g\dfrac {t^{2}}{2}(1)$ So at any given time the potential energy of the ball will be: $U\left( t\right) =mgH=mg\left( H_{0}+v_{0y}t-\dfrac {gt^{2}}{2}\right) \left( 2\right) $ So at any given time potential energy difference $\Delta U=U\left( t\right) -U\left( 0\right) =mg\left( H_{0}+v_{oy}t-g\dfrac {t^{2}}{2}\right) -mgHo=mg\left( v_{0y}t-g\dfrac {t^{2}}{2}\right) $ $v_{oy}=24\dfrac {m}{s};m=1kg;t=6s;g=9,8\dfrac {m}{s^{2}}$ ( becouse $+y$ axis is written as $j$ So we get $\Delta U=mg\left( v_{0y}t-g\dfrac {t^{2}}{2}\right) \approx -32.4J$
