Question

A body is falling from a vertical height of 10 m pierces through a distance of 1 m in sand. Calculate the average retardation in sand. {g=accel. Due to gravity.}

Answer 1

Answer on Question #44164 – Physics – Mechanics, Kinematics, Dynamics

Question:

A body is falling from a vertical height of 10 m pierces through a distance of 1 m in sand. Calculate the average retardation in sand. {g=accel. Due to gravity.}

Answer:

The law of conservation of energy:

$$\frac{m v^{2}}{2} + 0 = 0 + m g h$$

where $h$ is height, $g$ is acceleration due to gravity.

Therefore, speed before collision with sand equals:

$$v = \sqrt{2 g h}$$

Uniform deceleration in sand ($d = 1\ m$):

$$d = \frac{v^{2}}{2 a}$$

The average retardation equals:

$$a = \frac{v^{2}}{2 d} = \frac{2 g h}{2 d} = \frac{h}{d} g = 10 g = 98.1 \frac{m}{s^{2}}$$

Answer: $98.1\ \frac{m}{s^{2}}$

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