## College Physics (4th Edition)

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.