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 \[{{u}_{x}}=u\cos \theta \] Given: \[\theta =45{}^\circ \] \[\therefore \] \[{{u}_{x}}=u\cos 45{}^\circ =\frac{u}{\sqrt{2}}\] 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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