Question

Question 2

Answer 1

Answer:

Approximately $73\; {\rm N}$, assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$, is constant during the entire descent.

Let $x$ denote the displacement of this ball and let $t$ denote the duration of the descent. The SUVAT equation $x = (1/2)\, a\, t^{2}$ would apply.

Rearrange this equation to find an expression for the acceleration, $a$, of this ball:

$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$.

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$.

Let $m$ denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is $a$, the net external force on this ball would be $m\, a$.

Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$, the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$.