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3-2.Motion in Plane
normal
A particle is projected at $60^o $ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is
A$\frac{K}{2}$
B$K$
Cશૂન્ય
D$\;\frac{K}{4}$
Solution
$\begin{array}{l}
Let\,u\,be\,velocity\,with\,which\,the\,particle\\
is\,thrown\,and\,m\,be\,mass\,of\,the\,particle.Then\\
K = \frac{1}{2}m{u^2}.\\
At\,the\,highest\,{\rm{point}}\,{\rm{the}}\,{\rm{velocity}}\,{\rm{is}}\,{\rm{u}}\,{\rm{cos}}\,{\rm{6}}{{\rm{0}}^ \circ }\\
(only\,the\,horizontal\,component\,remains,the\\
velocity\,component\,begin\,zero\,at\,the\,\\
top – most\,{\rm{ point)}}{\rm{.}}\,Thereforce\,kinetic\,energy\\
at\,the\,highest\,{\rm{ point}}{\rm{.}}\\
K' = \frac{1}{2}m{\left( {u\cos {{60}^ \circ }} \right)^2} = \frac{1}{2}m{u^2}{\cos ^2}{60^ \circ } = \frac{K}{4}\,\left[ {From\,1} \right]
\end{array}$
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\,{\rm{point}}\,{\rm{the}}\,{\rm{velocity}}\,{\rm{is}}\,{\rm{u}}\,{\rm{cos}}\,{\rm{6}}{{\rm{0}}^ \circ }\\
(only\,the\,horizontal\,component\,remains,the\\
velocity\,component\,begin\,zero\,at\,the\,\\
top – most\,{\rm{ point)}}{\rm{.}}\,Thereforce\,kinetic\,energy\\
at\,the\,highest\,{\rm{ point}}{\rm{.}}\\
K' = \frac{1}{2}m{\left( {u\cos {{60}^ \circ }} \right)^2} = \frac{1}{2}m{u^2}{\cos ^2}{60^ \circ } = \frac{K}{4}\,\left[ {From\,1} \right]
\end{array}$
Standard 11
Physics
