Answer
$F = 6400~N$
Work Step by Step
We can find the speed when the person left the cannon:
$x = \frac{v^2~sin~2\theta}{g}$
$v^2 = \frac{x~g}{sin~2\theta}$
$v = \sqrt{\frac{x~g}{sin~2\theta}}$
$v = \sqrt{\frac{(69~m)(9.8~m/s^2)}{sin~(2)(53^{\circ})}}$
$v = 26.52~m/s$
We can find the acceleration in the barrel:
$v^2 = v_0^2+2ad$
$v^2 = 2ad$
$a = \frac{v^2}{2d}$
$a = \frac{(26.52~m/s)^2}{(2)(5.2~m)}$
$a = 67.63~m/s^2$
We can find the force $F$ propelling him:
$F - mg~sin~\theta = ma$
$F = m~(a+g~sin~\theta )$
$F = (85~kg)~[(67.63~m/s^2)+(9.8~m/s^2)~(sin~53^{\circ})]$
$F = 6400~N$
