Question

A 70 kg person lands on firm ground after jumping from a height of 3 m. What is the average force (in N) exerted on her feet by the ground if she bends her legs as she lands so that her body moves 50 cm during the impact? In other words, the change in her velocity as she comes to a stop occurs over a distance of 50 cm. You may assume that her acceleration is constant over the 50 cm.

Answer 1

4816N

Step-by-step explanation:

Height = 3m

Mass = 70kg

g = 9.8m/s²

S = 50cm = 0.5m

The total force = the weight of the body + impact force.

Impact force = ma but a = v / t

Using equation of motion,

V² = u² + 2as

V² = u² + 2gh

But u = 0m/s (since the body is at rest)

V² = 0 + 2 * 9.8 * 3

V² = 58.8

V = √(58.8)

v = 7.668m/s ≈ 7.67m/s

The impact interval = ?

Average velocity = total distance / time taken

V = 2s / T

T = (2 * s) / v

T = (2 * 0.5) / 7.67

T = 1 / 7.67

T = 0.13 seconds

The total force F = (mg + ma)

But a = v / t

F = m(g + v / t )

F = 70 * [ 9.8 + (7.67/0.13)]

F = 70 * (9.8 + 59)

F = 4816N