College Physics (4th Edition)

(a) We can find the vertical position of the ball after $1.60~s$: $y = y_0+\frac{1}{2}at^2$ $y = 14.0~m+\frac{1}{2}(-9.80~m/s^2)(1.60~s)^2$ $y = 1.46~m$ We can find the horizontal position after $1.60~s$: $x = v_x~t$ $x = (20.0~m/s)(1.60~s)$ $x = 32.0~m$ After 1.60 seconds have elapsed, the ball is 1.46 meters above the ground a horizontal distance of 32.0 meters away. (b) We can find the time it takes for the ball to hit the ground from a height of $14.0~m$: $\Delta y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2\Delta y}{a}}$ $t = \sqrt{\frac{(2)(-14.0~m)}{-9.80~m/s^2}}$ $t = 1.69~s$ We can find the horizontal position after $1.69~s$: $x = v_x~t$ $x = (20.0~m/s)(1.69~s)$ $x = 33.8~m$ The ball will hit the ground after another 0.09 seconds have elapsed. At this time, the ball will be a horizontal distance of 33.8 meters away from the starting point.