Question 186: 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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