Answer:
Approximately [tex]9.3\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Apply unit conversion:
[tex]t = 45\; {\rm ms} = 45 \times 10^{-3}\; {\rm s}[/tex].
At a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of an object of mass [tex]m[/tex] would be [tex]p = m\, v[/tex].
Initial momentum of this ball:
[tex]\begin{aligned}p_{0} &= m\, v_{0} \\ &= 0.060\; {\rm kg} \times (-32\; {\rm m\cdot s^{-1}}) \\ &= (-1.92\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}[/tex].
When a constant force [tex]F[/tex] is exerted on an object for a duration of length [tex]t[/tex], the impulse [tex]J[/tex] applied to that object would be [tex]J = F\, t[/tex].
Impulse that the ground applied to this ball:
[tex]\begin{aligned}J &= F\, t \\ &= 55\; {\rm N} \times (45 \times 10^{-3}\; {\rm s}) \\ &= 2.475\; {\rm N \cdot s}\end{aligned}[/tex].
Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex]. Thus, the impulse applied to this ball would be equivalent to:
[tex]\begin{aligned}J &= 2.475\; {\rm (kg \cdot m \cdot s^{-2}) \cdot s} \\ &= 2.475\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
After this impulse was applied, the momentum of this ball would become:
[tex]\begin{aligned}p_{1} &= p_{0} + J \\ &= (-1.92\; {\rm kg \cdot m \cdot s^{-1}}) + 2.475\; {\rm kg \cdot m \cdot s^{-1}} \\ &= 0.555\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The new velocity of this ball would be:
[tex]\begin{aligned}v_{1} &= \frac{p_{1}}{m} \\ &= \frac{0.555\; {\rm kg \cdot m \cdot s^{-1}}}{0.060\; {\rm kg}} \\ &\approx 9.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
