A) 1 : 1
B) 2 : 1
C) 3 : 2
D) 4 : 1
Correct Answer: D
Solution :
For first ball At highest point, the kinetic energy is completely converted into its potential energy \[\therefore \] \[mg{{h}_{1}}=\frac{1}{2}m{{u}^{2}}\] or \[{{h}_{1}}=\frac{{{u}^{2}}}{2g}\] For second ball \[mg{{h}_{2}}=\frac{mg{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[=\frac{1}{2}m{{u}^{2}}{{\sin }^{2}}{{30}^{o}}\] \[=\frac{1}{2}m{{u}^{2}}{{\left( \frac{1}{2} \right)}^{2}}\] or \[{{h}_{2}}=\frac{{{u}^{2}}}{8g}\] \[\therefore \] \[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{{{u}^{2}}}{2g}\times \frac{8g}{{{u}^{2}}}\] \[\Rightarrow \] \[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{4}{1}\]
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