A) \[\sqrt{3}:1\]
B) \[3:1\]
C) \[3:2\]
D) \[3:2\]
Correct Answer: D
Solution :
Let the velocity of two balls at the projection time is \[{{u}_{1}}\] and \[{{u}_{2}}\] and H be the corresponding height attained by both balls. Height of first ball = Highest of second ball \[\frac{u_{1}^{2}\,\,{{\sin }^{2}}\,{{45}^{o}}}{2g}=\frac{u_{2}^{2}\,\,{{\sin }^{2}}\,\,{{60}^{o}}}{2g}\] \[\Rightarrow \] \[u_{1}^{2}\times 1/2=u_{2}^{2}\times 3/4\] \[\Rightarrow \] \[{{u}_{1}}:{{u}_{2}}=\sqrt{3}:\sqrt{2}\]
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