Hold on, Indy!

Solution using energy:
The total distance that the heroine falls once the bridge hits the wall is 20 meters since the heroine is standing 2/3 the way across a 30 meter bridge.
We know that after the fall, all of the potential energy after 20 meters will be converted to kinetic energy.

$mv^{2} /2 = mgh$ via conservation of energy.
$v^{2}= 2gh$ * after cancelling the mass. Note: mass is irrelevant in this problem.
* $v=\sqrt{2 \times 9.8meters/second^{2} \times 20 meters}$ after taking the square root of both sides
The final answer for velocity comes out to 19.8 meters/second .

Moment of inertia of the bridge-heroine system about the point from where the bridge is hanging = (m1l^2)/3 + m2r^2 I = (500 30 30)/3 + 70 20 20 I = 178000 kgm^2 Work done by gravitational Torque dW = Ƭ sinө dө Total work done W = ʃ(0toπ/2) Ƭ sinө dө = ʃ(0toπ/2)(m1gr1 + m2gr2)sinө dө W = [(20 70 9.8)+(15 500 9.8)] (0-(-1)) = 87220 By work energy theorem W = ΔK.E. ½ Iω^2 = 87220 So; ω = 0.99 rad/s Velocity of heroine = r2 ω So; v = 20*0.99 v = 19.8 m/s

In this system, the total energy is conserved. As a result, $M_{bridge}gh_{cm,bridge} + M_{person}gh_{person} = \frac{1}{2}(I_{bridge}+M_{person}(\frac{2}{3}L)^{2})\omega^{2}$ .

$\omega$ is the final angular velocity of the bridge and the person when they reach the canyon wall.

$h_{cm,bridge}$ is the center of mass height above the final center of mass position of the bridge.

$h_{cm,person}$ is the height of the person above the final position of the person.

$I_{bridge}$ is the moment of inertia of the bridge about an axis at its end.

Using this equation, you can solve for $\omega$ to be approximately $0.990 rad/sec.$ .

The tangential velocity of the person would then be $\omega(\frac{2}{3}L) = \boxed{19.809 m/s}$ .

Let's isolate the bridge from the hero. First, calculate the bridge's final velocity: $PE_i + KE_i = PE_f + KE_f$ $mgh = \frac{mv^2}{2}$ $2gh = v^2$ The bridge has uniform velocity, so we can treat it as a point of mass located in the center of the bridge. $v^2 = 2gh = 2 \times g \times 15m = g \times 30m$ The same way, we find the hero's velocity squared to be $g \times 40m$ . They both have the same angular velocity, so that means one does not depend on the other. The hero's velocity is then: $\sqrt{9.8m/s^2 \times 40m} \approx 19.79m/s$

Because potential and kinetic energy are conserved, $PE_i = KE_f$ . Or, in other words, $mgh = \frac{mv^2}{2}$ and $v=\sqrt{gh}$ . In this case, since he is $\frac{2}{3}$ of the way, the radius - or height - is $\frac{2}{3}\left(30m\right) = 20m$ . Thus, $v=\sqrt{9.81\left(20\right)} = 19.809$ .

This is a relatively simple problem requiring just excessive calculations. Let mass of Bridge be M and heroine be m. Let she cling at a spot r from the center of rotation of the bridge. As length of bridge is l=30 m. I am taking reference level for calculation of P.E from that point Initial Mechanical Energy= Mg 30 +mg 30. Final M. E= Mg 15 + mg (30-2/3*30) +1/2 I Omega^2....the last tem is due to M.I. Equate both sides and the asnwer is obtained...