Answer:
Approximately [tex]4.2 \times 10^{2}\; \rm W[/tex] on average in the vertical direction (assuming that [tex]g = 9.8\; \rm N \cdot kg^{-1}[/tex].)
Explanation:
Vertical displacement of this student in [tex]6\; \rm s[/tex]: [tex]36 \times 16\; \rm cm = 36 \times (0.16\; \rm m) = 5.76\; \rm m[/tex].
Average vertical velocity of this student: [tex]v = \displaystyle \frac{5.76\; \rm m}{6\; \rm s} = 0.96\; \rm m \cdot s^{-1}[/tex].
The average vertical force that this student exerts on the ground would be equal to the weight of this student:
[tex]F = W = m \cdot g = 45\; \rm kg \times 9.8\; \rm N \cdot kg^{-1} = 441\; \rm N[/tex].
The average power of this student in the vertical direction is equal to the product of the force and velocity of this student in that direction:
[tex]P = F \cdot v = 441\; \rm N \times 0.96\; \rm m \cdot s^{-1} \approx 4.2\times 10^{2}\; \rm W[/tex].
