Answer
The projectile would have gone $~~738~m~~$ higher.
Work Step by Step
We can find the height difference associated with a gravitational potential energy of $68.0~kJ$:
$mg~\Delta h = 68.0~kJ$
$\Delta h = \frac{68.0~kJ}{mg}$
$\Delta h = \frac{68.0~kJ}{(9.40~kg)(9.80~m/s^2)}$
$\Delta h = 738~m$
The projectile would have gone $~~738~m~~$ higher.
