Answer
After 1.40 seconds have elapsed, the ball is on the ground a horizontal distance of 42.0 meters away.
Work Step by Step
We can find the vertical position after $1.40~s$:
$y = y_0+\frac{1}{2}at^2$
$y = 9.60~m+\frac{1}{2}(-9.80~m/s^2)(1.40~s)^2$
$y = 0~m$
We can find the horizontal position after $1.40~s$:
$x = v_x~t$
$x = (30~m/s)(1.40~s)$
$x = 42.0~m$
After 1.40 seconds have elapsed, the ball is on the ground a horizontal distance of 42.0 meters away.
