Question

A metal ball of mass 0.5kg is dropped from top of a vertical Clift of height 90m. When it hits the beach below it penetrates to a depth of 6.0cn calculate the average retarding force of the sand (neglect air resistance) I know the answer to this question is 7.5kN but I don’t understand how you get to answer

Answer 1

Answer on Question #72006 Physics / Other

A metal ball of mass $m = 0.5 \, \mathrm{kg}$ is dropped from top of a vertical Clift of height $h = 90 \, \mathrm{m}$. When it hits the beach below it penetrates to a depth of $S = 6.0 \, \mathrm{cm}$ calculate the average retarding force of the sand (neglect air resistance).

Solution:

The velocity of the ball when it hits the beach

$$v_i = \sqrt{2gh}$$

After penetration in the sand the ball stopped. So final velocity

$$v_f = 0$$

The depth of preparation

$$S = \frac{v_i^2 - v_f^2}{2a} = \frac{v_i^2}{2a} = \frac{gh}{a}$$

Where $a$ is acceleration of the ball due to retarding force of the sand.

The Newton's second law

$$F = ma$$

gives

$$F = m \frac{gh}{S}$$ $$F = \frac{0.5 \times 10 \times 90}{0.06} = 7500 \, \mathrm{N} = 7.5 \, \mathrm{kN}.$$

Answer: $F = 7.5 \, \mathrm{kN}$.