Answer
The ball slides a distance of $~~8.7~m$
Work Step by Step
In part (b), we found that the linear acceleration is $-2.1~m/s^2$
In part (d), we found that the ball slides for a time of $1.2~s$
We can find the distance that the ball slides:
$x = v_0~t+\frac{1}{2}at^2$
$x = (8.5~m/s)(1.2~s)+\frac{1}{2}(-2.1~m/s^2)(1.2~s)^2$
$x = 8.7~m$
The ball slides a distance of $~~8.7~m$
