Calculate the impulse experienced when a 74kg person lands on firm ground after

Question

Calculate the impulse experienced when a 74kg person lands on firm ground after jumping from a height of 2.9m .

Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged. With stiff legs, assume the body moves 2.0cm during impact. [Hint: The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.]

Estimate the average force exerted on the person's feet by the ground if the landing is with bent legs. When the legs are bent, assume the body moves about 46cm during impact.

For the third question, my answer is 4600N, but it is wrong. Please help me with this

Explanation / Answer

The average force is the rate of change of momentum.

Let:
u be the landing speed,
h be the height of the jump,
g be the acceleration due to gravity,
P be the momentum on landing,
I be the impulse,
s be the stopping distance after landing.

u^2 = 2gh
u = sqrt(2gh)
= 7.5392 m/s.

P = mu
= m sqrt(2gh)
= 557.903 kg m/s
= 558 m/s to 3sf.

s = t(u + 0) / 2
t = 2s / u

F = mu / t
= P / t
= Pu / (2s)

(b)
F = 557.903 * 7.5392 / (2 * 0.02)
= 1.05153 * 10^5 N to 3sf.

(c)
F = 557.903 * 7.5392 / (2 * 0.46)
= 4571.893 N to 3sf.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.