Answer:
Approximately [tex]2.9 \; {\rm kg\cdot m\cdot s^{-1}}[/tex].
Explanation:
Under the assumptions, the baseball will accelerate downwards at a constant acceleration [tex]a[/tex] of [tex]a = g = 9.8\; {\rm m\cdot s^{-2}}[/tex].
Let [tex]x[/tex] denote the displacement of the baseball. Let [tex]u[/tex] denote the initial velocity of the baseball. Let [tex]v[/tex] denote the velocity of the baseball right before landing.
It is given that the baseball in this question was initially at rest, such that [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]. It is also known that the displacement was [tex]x = 22\; {\rm m}[/tex]. The SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] relates these quantities.
Rearrange this equation to find the velocity of the baseball right before landing:
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \, (9.8\; {\rm m\cdot s^{-2}})\, (22\; {\rm m}) + (0\; {\rm m\cdot s^{-1}})^{2}} \\ &\approx 20.77\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
When an object of mass [tex]m[/tex] travels with a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].
Thus, the momentum of this baseball right before landing will be:
[tex]\begin{aligned}p &= m\, v \\ &\approx (0.14\; {\rm kg})\, (20.77\; {\rm m\cdot s^{-1}}) \\ &\approx 2.9\; {\rm kg\cdot m\cdot s^{-1}}\end{aligned}[/tex].
