A) zero
B) E/2
C) \[E/\sqrt{2}\]
D) E
Correct Answer: B
Solution :
Key Idea: At the highest point, only horizontal velocity is present. At highest point vertical component of velocity is zero; while horizontal component remains unchanged. Velocity at highest point, \[{{v}_{x}}=v\,\cos {{45}^{o}}\] and \[{{v}_{y}}=0\] \[\therefore \] Kinetic energy at highest point \[E=\frac{1}{2}mv_{x}^{2}\] \[=\frac{1}{2}m{{(v\,\cos \,{{45}^{o}})}^{2}}\]c \[=\frac{1}{2}m{{v}^{2}}\,\,{{\cos }^{2}}\,{{45}^{o}}\] \[=\frac{1}{2}m{{v}^{2}}\,\times {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\] \[=\frac{1}{4}m{{v}^{2}}\,\] Initial kinetic energy, \[E=\frac{1}{2}m{{v}^{2}}\] \[\therefore \] \[E=\frac{1}{4}m{{v}^{2}}\] \[=\frac{1}{2}\left( \frac{1}{2}m{{v}^{2}} \right)=\frac{1}{2}E\]
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