Question

Bungee jumping: a man falls from the tower H = 200 m height with a rubber tope attached to his feet. The minimal height down to which the jumper will descend is h′ = 50 m over the ground. After all the oscillation will stop, the jumper will hang at the height h = 90 m over the ground. What maximal speed will the jumper reach during this jump? The air resistance may be neglected. g = 10 м/с2
Please, express result is m/s and write it down rounded
up to 3 significant digits

Answer 1

The maximal speed will be reached in the point 1 at the height h.. This is the equilibrium point where the gravity forced is compensated by the elastic force. After some oscillations the jumper will stop exactly at this point. If l₀ is the length of the free bungee (not expanded, nor compressed)

The energy conservation law for the points 0 and 1 at height h gives:

$mgH =mgh+1/2 k( H- h- l_0)^2 +1/2mv^2$ ........(1)

(k- is the elasticity coefficient of the bungee).

For the points 0 and 2 (at the height h') we obtain:

$mgH =mgh+1/2 k( H- h- l_0)^2$ ................(2)

Equilibrium condition at the point 1:

$mg=k(H-h-l_0)$ ......................(3)

Substituting (3) into (2) we obtain

$l_0^2=2(H-h')(H-h)-(H-h)^2$

putting the values we get

$l_0^2=102m$

Substituting k into (1), we may additionally find the maximal velocity of the jumper

$v=(g(H-h+l_0))^{1/2}$ =46.1m/s