Answer:
No. of days, n = 0.68 days
Solution:
As per the question:
Energy of one day diet, [tex]E_{d} = 2200\ large\ calories[/tex]
Height above sea level, H = 8848 m
Mass, M = 72 kg
Now,
1 large calorie = [tex]4.2\times 10^{3} J[/tex]
Therefore,
[tex]E_{d} = 2200\times 4.2\times 10^{3} = 9.24\times 10^{6}J[/tex]
Now,
The potential energy required for jumping of the Mt. Everest at a height of 8848 m:
[tex]PE_{g} = MgH = 72\times 9.8\times 8848 = 6.24\times 10^{6} J[/tex]
No. of days can be given as:
[tex]n = \frac{PE_{g}}{E_{d}}[/tex]
[tex]n = \frac{6.24\times 10^{6}}{9.24\times 10^{6}} = 0.68\ days[/tex]
