Question

explain the total mechanical energy of a free falling object always remain constant

Answer 1

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Suppose We Take A Body To Height h above the ground level

Then It Must be at rest hence it must possess Potential Energy = mgh

Now We Drop The Body And It Covers a Distance x From Its Initial Position

Then height from ground = h-x

At this moment it must both have PE AND KE

PE = mg(h-x) = mgh-mgx

KE = 1/2*m*v2 1)

Now Using Third Equation OF motion

v2 = u2+2as

here u = 0

v2 = 2*g*x

v2 = 2gx

put above value in equation 1

KE = 1/2*m*2gx

KE = mgx

Now PE+KE = mgh-mgx+mgx = mgh

Now when it hits the ground

h = 0

then it must only have kinetic energy

KE = 1/2*m*v2

v2 = u2+2as

v2 = 2gh

put this value in KE Equation

KE = 1/2*m*2gh = mgh

Hence Its Total Energy Remains Constant

Hope it'll help
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