Answer
(a) After 1.60 seconds have elapsed, the ball is 1.46 meters above the ground a horizontal distance of 32.0 meters away.
(b) 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.
Work Step by Step
(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.
