A) \[+\frac{E}{\sqrt{2}}\]
B) \[+\frac{E}{2}\]
C) \[-\frac{E}{\sqrt{2}}\]
D) \[-\frac{E}{2}\]
Correct Answer: B
Solution :
: At starting point, potential energy = 0 Maximum height attained by projectile\[=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\therefore \]\[h=\frac{{{u}^{2}}{{\sin }^{2}}45{}^\circ }{2g}\] or \[h=\frac{{{u}^{2}}}{4g}\] \[\therefore \]Potential energy gained\[=mgh\] \[PE=\frac{m{{u}^{2}}}{4}\] \[PE=\frac{1}{2}\times \frac{1}{2}m{{u}^{2}}=\frac{1}{2}\times E\] or PE gained\[=E/2\].
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