For a body of mass m kg, show that the depth to which the body would fall if attached to a rope, with a length of l meters. The depth is given by the model:
$$d= \frac{2ml \pm l \cdot \sqrt{4m^2 + 600m}}{150} + l$$
The rope is specially designed and its modulus of elasticity is known from specifications.
For the purposes of this problem, assume that the rope is stretched to twice its natural length by a body of 75kg hanging at rest from the free end.
Define one assumption before beginning the problem.
I know that if I rearrange the equation I get, $75(d-l)^2 - 2ml(d-l) + 2ml = 0$. This indicated to me that the problem has probably come from a kinematic equation. However, once I reach this point I am stuck. I don't know how to progress any further into the question. Would simple harmonic motion equations be required? Thanks in advance.
The energy approach is like this:
The elastic potential energy is given by $$EPE=\frac {\lambda x^2}{2l}$$ where $\lambda$ Is the modulus of elasticicty, in this case $75g$, $x$ is the extension, and $l$ is the natural length.
The particle starts from rest and falls, coming instantaneously to rest after falling a distance $d$. Assuming no air resistance, total initial energy $=$ total final energy, and kinetic energy is zero initially and finally, so we have: $$0=\frac{75g(d-l)^2}{2l}-mgd$$
Can you take it from there?