A) \[\frac{K}{2}\]
B) \[\frac{K}{\sqrt{2}}\]
C) K
D) zero
Correct Answer: A
Solution :
Key Idea: At highest point of projection, the vertical component of velocity is zero and there is only horizontal component of velocity. At the highest point \[{{v}_{x}}=u\cos \theta \] \[{{v}_{y}}=0\] \[{{K}_{H}}=\frac{1}{2}mv_{x}^{2}\] \[or{{K}_{H}}=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta .....(i)\] Initial kinetic energy is \[K=\frac{1}{2}m{{u}^{2}}...(ii)\] From Eq. (i) and (ii), we get \[{{K}_{H}}=K{{\cos }^{2}}\theta =K{{\cos }^{2}}{{45}^{o}}\] \[=K\times {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}=\frac{K}{2}\]
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