# Question #34e08

Jan 6, 2018

The distance the ball can reach is $\approx 5.21 m$

#### Explanation:

The initial speed of ball is

$u = 2.5 m {s}^{-} 1$

The ball is decelerating due to friction

$a = - 0.6 m {s}^{-} 2$.

When it comes to rest after distance $s$, the final velocity will be

$v = 0 m {s}^{-} 1$

According to the position- velocity relation:

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

$s = \frac{0 - {2.5}^{2}}{2 \left(- 0.6\right)}$ ${\left(m {s}^{-} 1\right)}^{2} / \left(m {s}^{-} 2\right)$

$s = - \frac{6.25}{-} 1.2$ $\frac{{m}^{2} {\cancel{s}}^{-} 2}{m {\cancel{s}}^{-} 2}$

$s = 5.208 \overline{33}$ $m$

$\therefore$ The distance the ball can reach is $\approx 5.21 m$