So,
\(\frac{1}{2}m{v^2} = mgh;\;\)
\(Here\;h = 1\;m\)
\(v = \sqrt {2g} \)
Now Maxium power delivered by musheles is given by
\(P\; = 2Fv = 2 \times mg \times \sqrt {2g} \)
\( = 2 \times 70 \times 10 \times \sqrt {20} \)
\( = 6260.8 = 6.26 \times {10^3}\;W/s\)