A bungee jumper is jumping with the help of a | vedclass

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(D) $1$. For the jumper to come to rest at distance $S$,the loss in gravitational potential energy equals the gain in elastic potential energy: $mgS =

Vedclass · Accepted Answer

Time to come to rest for the first time $= \left( \sqrt{\frac{2l}{g}} + \sqrt{\frac{m}{K}} \left( \frac{\pi}{2} + \arcsin\left( \frac{mg}{mg + Kl} \right) \right) \right)$

Vedclass · Answer

(D) $1$. For the jumper to come to rest at distance $S$,the loss in gravitational potential energy equals the gain in elastic potential energy: $mgS = \frac{1}{2}K(S-l)^2$. Solving this quadratic equation $KS^2 - 2(Kl+mg)S + Kl^2 = 0$ gives $S = \frac{(Kl+mg) + \sqrt{2mgKl + m^2g^2}}{K}$. Thus,option $A$ is correct.$2$. Maximum speed occurs at the equilibrium position where the net force is zero: $mg = Kx_{eq} \Rightarrow x_{eq} = \frac{mg}{K}$. The total distance fallen is $l + \frac{mg}{K}$. By conservation of energy: $mg(l + \frac{mg}{K}) = \frac{1}{2}mv^2 + \frac{1}{2}K(\frac{mg}{K})^2$. Solving for $v$ gives $v = \sqrt{2gl + \frac{mg^2}{K}}$. Thus,option $B$ is correct.$3$. The time of free fall for distance $l$ is $t_1 = \sqrt{\frac{2l}{g}}$. Thus,option $C$ is correct.$4$. The motion after the rope becomes taut is a simple harmonic motion starting from the natural length position with an initial downward velocity $v_0 = \sqrt{2gl}$. The time to reach the lowest point is not simply $\frac{\pi}{2}\sqrt{\frac{m}{K}} + \sqrt{\frac{2l}{g}}$. Therefore,option $D$ is the incorrect statement.

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