# Physics Question?

The St. Louis Arch has a height of 192 m.

Suppose that a stunt woman of mass 96 kg

jumps off the top of the arch with an elastic

band attached to her feet. She reaches the

ground at zero speed.

The acceleration of gravity is 9.81 m/s^2.

Find her kinetic energy after 3.8 s of the

flight. Assume the elastic band has no length

and obeys Hooke’s Law.

Answer in units of kJ.

### 3 Answers

- NCSLv 72 years agoFavorite Answer
GPE becomes SPE

mgh = ½kh²

k = 2mg / h = 2*96*9.81 / 192 = 9.81 N/m

period T = 2π√(m/k) = 2π√(96 / 9.81) = 19.655 s

so we can model the motion with

y = h/2 + (h/2)cos(2πt/T)

where h/2 = amplitude = 96 m

and 2π/T = 2π/19.665s = 0.320 rad/s

so

y = 96 + 96cos(0.320t) → for y in meters and t in seconds

Then v = dy/dt = -96 * 0.320 * sin(0.320t) = -30.7sin(0.320t)

for v in m/s.

At t = 3.8 s, we have

v = -30.7*sin(0.320*3.8) = -28.8 m/s

(that is, down)

KE = ½mv² = ½ * 96kg * (-28.8m/s)² = 40 000 J = 40 kJ

to two significant digits (39 712 J before rounding).

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- oubaasLv 72 years ago
2*m*g*h = k*h^2

k = √2*96*g/192 = √g N/m

m*g*h = m*g*(h-Δh)+m/2*2*Δh*g+(√g)*(Δh)^2/2

2*m*g*(192-192+Δh-Δh) = √g*(Δh)^2

2*m*√g = (Δh)^2

Δh = √2*m*√g = √2*96*3.132 = 24.52 m

Ekx = m/2*V^2 = 46*2*g*Δh = 19.62*24.52*46/1000 = 22.130 kjoule

- Firstname2 years agoReport
Hey, oubaas. I've tried your answer and even calculated it to a few more significant figures, but it isn't showing up as correct on my online homework. Can you please re-examine it? Thanks in advance.

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- AmyLv 72 years ago
The elastic potential energy at 192m extension is equal to the gravitational potential energy at 192 m high.

1/2 k (192 m)^2 = (96 kg) (9.81 m/s^2) (192 m)

k = 9.81 kg/s^2

That means her acceleration is:

a = g - kx/m = 9.81 - 0.102 x m/s^2

where x is distance fallen in meters.

Solve the differential equation to relate distance to time.

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