Question

1. Let the force of gravity to be constant for small distances above the surface of the earth. A body is dropped from rest at a height h above the earth surface. What will be its kinetic energy just before it strikes the ground?   

Answer 1

Answer

A body is dropped from rest at a height h above the earth surface. its kinetic speed just before it strikes the ground is given

$V=\sqrt{2g(R_e+h) }$

Where R_{e is radius of earth.}

Now kinetic energy is given by

$KE=\frac{m\times2g(R_e+h) }{2}\\=m\times g(R_e+h)$