Question 185: Work, Energy and Power & Circular Motion

Question

A particle is projected at $$60^\circ $$ to the horizontal with a kinetic energy K. The kinetic energy at the
highest point is

Accepted Answer

Let $$u$$ be the velocity with which the particle is thrown and $$m$$ be the mass of the particle. Then
$$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
At the highest point the velocity is $$u$$ $$\cos \,{60^ \circ }$$ (only the horizontal component remains, the vertical component being zero at the top-most point).
Therefore kinetic energy at the highest point,
$${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ $$
$$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$$  [ From  eq $$(1)$$ ]

Answer

K/2

Answer

K

Answer

Zero

Answer

K/4

Work, Energy and Power & Circular Motion

A particle is projected at $60^\circ $ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

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