question_answer

A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of \[60{}^\circ \] with the vertical with the same initial speed. At highest points of their journey, the ratio of their potential energies will be

A) \[1:1\]                           

B) \[2:1\]

C) \[3:2\]               

D)        \[4:1\]

Correct Answer: D

Solution :

For first ball, \[mg{{h}_{1}}=\frac{1}{2}m{{\nu }^{2}}\] \[{{h}_{1}}=\frac{{{u}^{2}}}{2g}\] For second ball \[mg{{h}_{2}}=mg\frac{{{u}^{2}}\cos \,\theta }{2\,g}\] \[=\,\,\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta =\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}60{}^\circ \] \[=\,\,\frac{1}{2}m{{u}^{2}}{{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\,m{{u}^{2}}\,\left( \frac{1}{4} \right)\] \[\Rightarrow \,\,\,{{h}_{2}}=\frac{{{u}^{2}}}{8g}\] \[\therefore \,\,\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{{{u}^{2}}}{2g}\times \frac{8g}{{{u}^{2}}}\,\,\Rightarrow \,\,\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{4}{1}\]