A 70 kg man leaps vertically into the air from a crouching position. To take the leap the man pushes the ground with a constant force F to raise himself. The center of gravity rises by 0.5 m before he leaps. After the leap the c.g. rises by another 1m. the maximum power delivered by the muscles is (Take g=10 ms^–2)

Mass of man m = 70kg

Man pushes the ground with force F to take the leap and his center of gravity rises by 1m after the leap.

Law of conservation of energy is applied .

Kinetic energy initially = potential energy finally

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh} \\ \Rightarrow \mathrm{v}=\sqrt{2\mathrm{gh}} \end{gathered}$$

$\mathrm{h}=1\Rightarrow \mathrm{v}=\sqrt{2\times 10\times 1}=\sqrt{20}m/s$

The maximum power delivered by the muscles

$\mathrm{P}=2\times \mathrm{mg}\times \mathrm{v}=2\times 70\times 10\times \sqrt{20}$

$\mathrm{P}=6.26\times {10}^3 \mathrm{Watts} \mathrm{at} \mathrm{take} \mathrm{off}.$

Hence the correct answer is $6.26\times {10}^3\mathrm{Watts} \mathrm{at} \mathrm{take} \mathrm{off}.$