Answer:
In case of a projectile, the
potential energy is maximum at the highest point, given by
\[\therefore \] \[{{\left(
P.E. \right)}_{H}}=mgH=mg\frac{{{u}^{2}}{{\sin }^{2}}\theta
}{2g}=\frac{1}{2}m{{u}^{2}}{{\sin }^{2}}\theta \]
Kinetic energy of
projectile will be minimum (but not zero) at the highest point, because only
vertical component velocity is zero there. Therefore \[{{\left( K.E.
\right)}_{H}}=\frac{1}{2}mu_{H}^{2}=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta \]
\[\therefore \] \[{{\left(
P.E. \right)}_{H}}+{{\left( K.E. \right)}_{H}}=\frac{1}{2}m{{u}^{2}}\left(
{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=\frac{1}{2}m{{u}^{2}}\]
which is the total
mechanical energy of projectile at the point of projection. So in projectile
motion, total mechanical energy is conserved i.e. same at all locations of its
motion. K.E. of projectile is maximum at place of projection.
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