Question

A body of mass 3 kg dropped the top a tower h = 25 the k.E after 3 sec will be

Answer 1

Answer:

The initial K.E. of the body =12mu2 and it falls through a vertical height h.

It loses its P.E.=mgh and it is converted into K.E.

Explanation:

If v is the velocity with which the body strikes the ground, then its K.E when it hits the ground. =12mv2 and by the principle of conservation of energy 12mv2=12μ2+mgh

Thus v does not depend upon the direction in which it is thrown or it is independent of the angle of projection.