Question

show that the total energy of a freely falling bady remains constant during its motion ​

Answer 1

Answer:

Let the energy of particle at any height = E

Initial energy =

2

1

mv

2

+mgH

=0+mgH=E

o

Velocity at any height = v

∴v

2

=u

2

+2as

v

2

=0+2(−g)[−(H−h)]

=2g(H−h)

∴E=

2

1

mv

2

+mgh

=

2

m

.2g(H−h)+mgh=mgH=E

o

Answer 2

Answer:

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