Answer
The horizontal distance would have been $~~27~m~~$ greater.
Work Step by Step
We can find the time of the ball's flight:
$y = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$0 = (30~m)+(82~m/s)(sin~45^{\circ})~t+\frac{1}{2}(-9.8~m/s^2)~t^2$
$4.9~t^2-58~t-30 = 0$
We can use the quadratic formula to find $t$:
$t = \frac{58\pm \sqrt{(-58)^2-(4)(4.9)(-30)}}{(2)(4.9)}$
$t = \frac{58\pm \sqrt{3952}}{9.8}$
$t = -0.50~s, 12.3~s$
We can choose the positive solution.
We can find the horizontal displacement:
$x = v_x~t$
$x = (82~m/s)(cos~45^{\circ})(12.3~s)$
$x = 713~m$
We can find the difference in horizontal displacement:
$(713~m)-(686~m) = 27~m$
The horizontal distance would have been $~~27~m~~$ greater.
