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Question

$\begin{array}{l}
Let\,u\,be\,velocity\,with\,which\,the\,particle\
is\,thrown\,and\,m\,be\,mass\,of\,the\,particle.Then\
K = \frac{1}{2}m{u^2}.\

Vedclass · Accepted Answer

$\;\frac{K}{4}$

Vedclass · Answer

$\begin{array}{l}
Let\,u\,be\,velocity\,with\,which\,the\,particle\
is\,thrown\,and\,m\,be\,mass\,of\,the\,particle.Then\
K = \frac{1}{2}m{u^2}.\
At\,the\,highest\,{m{point}}\,{m{the}}\,{m{velocity}}\,{m{is}}\,{m{u}}\,{m{cos}}\,{m{6}}{{m{0}}^ \circ }\
(only\,the\,horizontal\,component\,remains,the\
velocity\,component\,begin\,zero\,at\,the\,\
top - most\,{m{ point)}}{m{.}}\,Thereforce\,kinetic\,energy\
at\,the\,highest\,{m{ point}}{m{.}}\
K' = \frac{1}{2}m{\left( {u\cos {{60}^ \circ }} ight)^2} = \frac{1}{2}m{u^2}{\cos ^2}{60^ \circ } = \frac{K}{4}\,\left[ {From\,1} ight]
\end{array}$

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

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