Answer:
b) √[(kx²/m) - 2gx]
Explanation:
The energy at the lowest point is equal to:
[tex]E_{elas}=\frac{1}{2} *k*x^{2}[/tex]
where:
Eelas = elastic energy [J]
k = spring constant [N/m]
x = extension of the spring [m]
We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.
[tex]E_{elas}=E_{pot}+E_{kine}\[/tex]
[tex]\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2} \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}[/tex]
