Answer:
Approximately [tex]1\; \rm kg[/tex]. (Assumption: [tex]g = 9.8\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Let the mass of this ball be [tex]x\; \rm kg[/tex].
Initial gravitational potential energy of this ball: [tex]m \cdot g \cdot h \approx (39.2\, x) \; \rm J[/tex].
Just before the ball hits the ground, all [tex](39.2\, x)\; \rm J[/tex] of gravitational potential energy would have been converted to kinetic energy. Calculate the velocity of the ball at that moment:
[tex]\begin{aligned} \text{kinetic energy} = \frac{1}{2}\, m \cdot v^{2} \end{aligned}[/tex].
Therefore, right before hitting the ground, the velocity of the ball would be:
[tex]\begin{aligned} v&= \sqrt{\frac{2\, (\text{kinetic energy})}{m}} \\ &= \sqrt{\frac{2 \times (39.2\, x)}{x}}\\ & \approx \sqrt{2 \times 39.2} \approx 8.85\; \rm m \cdot s^{-1}\end{aligned}[/tex].
The momentum [tex]p[/tex] of an object of mass [tex]m[/tex] and velocity [tex]v[/tex] would be [tex]p = m \cdot v[/tex]. Rewrite this equation to find an expression for mass [tex]m\![/tex] given velocity [tex]v\![/tex] and momentum [tex]p\![/tex]:
[tex]\displaystyle m = \frac{p}{v}[/tex].
Right before collision, the momentum of this ball is [tex]p = 8.85\; \rm kg \cdot m \cdot s^{-1}[/tex] while its velocity is [tex]v \approx 8.85\; \rm m \cdot s^{-1}[/tex]. Therefore, the mass of this ball would be:
[tex]\begin{aligned}m &= \frac{p(\text{right before landing})}{v(\text{right before landing})} \\ &\approx \frac{8.85\; \rm kg \cdot m \cdot s^{-1}}{8.85\; \rm m \cdot s^{-1}} \approx 1\; \rm kg\end{aligned}[/tex].
