A body of mass m = 2 kg is thrown downward from a height H =…

A body of mass $m = 2\ \mathrm{kg}$ is thrown downward from a height $H = 200\ \mathrm{m}$ at $v_0 = 198\ \mathrm{m/s}$. It strikes sand and sinks $h = 0.5\ \mathrm{m}$ into it. Find the average resistance force of the sand (neglect air resistance).

Explanation of Method

Average resistance of sand.

Step 1: Energy balance

The sand must remove the initial kinetic energy and all the potential energy released, down to the stopping point: $\langle F\rangle h = \tfrac12 mv_0^2 + mg\,(H + h)$.

\[ \langle F\rangle h = 39\,204 + 3933.8. \]

\[\langle F\rangle = \frac{43\,137.8}{0.5} \approx 86.3\ \mathrm{kN}\]

$v_0 = 198\ \mathrm{m/s}$ $H = 200\ \mathrm{m}$ $h = 0.5\ \mathrm{m}$ $\text{sand}$ $\langle F\rangle h = \tfrac12 mv_0^2 + mg(H + h)$

Not to scale. The sand's resistance has to remove the initial kinetic energy plus all the potential energy released, including the 0.5 m inside the sand.

Result

\[\boxed{\,\langle F\rangle = \frac{43\,137.8}{0.5} \approx 86.3\ \mathrm{kN}\,}\]

Remark

Check. The original result is correct (its $2 \cdot 9.81 \cdot 200.5$ should be 3933.8, not 3931.8, which does not change the answer). The force is about 4400 times the body's weight.