Question

The human tibia breaks under an impulse of 55 Ns. If after falling a person typically comes to eat in a time span of 0.005 s how fast would a 75 kg person need to be falling to break their leg?

Answer 1

0.73 m/s

Step-by-step explanation:

From Newton second law of motion,

I = m(v-u)...................... Equation 1

Where I = Impulse, m = mass of the person, v = final velocity, u = Initial velocity.

make v the subject of the equation

v =(I/m)+u................. Equation 2

Note: u = 0 m/s as the person is falling from an height.

Given: I = 55 Ns, m = 75 kg, u = 0 m/s

Substitute into equation 2

v = 55/75

v = 0.73 m/s