Answer:
Approximately [tex]73\; {\rm N}[/tex], assuming that the acceleration of this ball is constant during the descent.
Explanation:
Assume that the acceleration of this ball, [tex]a[/tex], is constant during the entire descent.
Let [tex]x[/tex] denote the displacement of this ball and let [tex]t[/tex] denote the duration of the descent. The SUVAT equation [tex]x = (1/2)\, a\, t^{2}[/tex] would apply.
Rearrange this equation to find an expression for the acceleration, [tex]a[/tex], of this ball:
[tex]\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}[/tex].
Note that [tex]x = 11\; {\rm m}[/tex] and [tex]t = 1.5\; {\rm s}[/tex] in this question. Thus:
[tex]\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is [tex]a[/tex], the net external force on this ball would be [tex]m\, a[/tex].
Since [tex]m = 7.5\; {\rm kg}[/tex] and [tex]a \approx 9.78\; {\rm m\cdot s^{-2}}[/tex], the net external force on this ball would be:
[tex]\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}[/tex].
