Answer
The pull of gravity changes the velocity of the ball by 0.51%.
Work Step by Step
$v_x = 150 ~km/h\times 1000m/km\times 1h/3600~s = 41.67 ~m/s$
$t = \frac{18 ~m}{41.67 ~m/s} = 0.432 ~s$
$v_y = v_{y,0} + at = 0 + 9.80\times 0.432 = 4.234 ~m/s$
$v = \sqrt{41.7^2 + 4.234^2} = 41.91 ~m/s$
The percentage difference is $\frac{41.91-41.7}{41.7}\times 100\% = 0.51\%$
