After scaling a wall of 3 m height a man of weight W drops himself to the ground. If his body comes to a complete stop 0.15 sec. After his feet touch the ground, calculate the average impulsive force in the vertical direction exerted by ground on his feet. (g = 9.8 m/s2)

Engineering

Physics

Acceleration

Straight Line Motion

Momentum and Energy Conservation for System of Particles

Question

After scaling a wall of 3 m height a man of weight W drops himself to the ground. If his body comes to a complete stop 0.15 sec. After his feet touch the ground, calculate the average impulsive force in the vertical direction exerted by ground on his feet. (g = 9.8 m/s²)

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To find the average impulsive force, we first determine the velocity when the man touches the ground. Using conservation of energy, the velocity v just before impact is given by $\frac{1}{2} m v^{2} = m g h$ . Solving, $v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 3} = 7.67 m / s$ downward.

The impulse-momentum theorem states that the average force F_avg times the time Δt equals the change in momentum: $F_{avg} Δ t = m Δ v$ . The change in vertical velocity Δv is 0 - (-7.67) = 7.67 m/s upward. Substituting, $F_{avg} \times 0.15 = m \times 7.67$ . Solving, $F_{avg} = \frac{m \times 7.67}{0.15} = 51.13 m$ . Since weight W = mg, m = W/g. Thus, $F_{avg} = 51.13 \frac{W}{g} = 51.13 \frac{W}{9.8} = 5.22 W$ .

Final Answer: The average impulsive force is 5.22W upward.