A) E
B) \[E/\sqrt{2}\].
C) \[E/2\]
D) zero
Correct Answer: C
Solution :
At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left. Which is as, there is no acceleration of particle in horizontal component A velocity remains constant \[{{u}_{x}}=u\cos \theta \] Given \[\theta ={{45}^{o}}\] \[\therefore \] \[{{u}_{x}}=u\cos {{45}^{o}}=\frac{u}{\sqrt{2}}\] Hence, at the highest point kinetic energy, \[E'=\frac{1}{2}mu_{x}^{2}=\frac{1}{2}m{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}=\frac{1}{2}m\left( \frac{{{u}^{2}}}{2} \right)\] \[=\frac{E}{2}\,\,\left( \because \frac{1}{2}m{{u}^{2}}=E \right)\]
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