question_answer

A body is projected with a speed \[\mu \,m/s\]at an angle \[\beta \]with the horizontal. The kinetic 3 energy at the highest point is \[\frac{3}{4}\,\text{th}\]of the 4 initial kinetic energy. The value of \[\beta \] is

A)  \[{{30}^{o}}\]                                   

B)  \[{{45}^{o}}\]

C)  \[{{60}^{o}}\]                                   

D)  \[{{120}^{o}}\]

Correct Answer: A

Solution :

                                The kinetic energy at the highest point would be equal to \[\frac{1}{2}m\,{{(u\,\cos \,\beta )}^{2}}\] as the vertical component of the velocity is zero. The initial kinetic energy is the maximum kinetic energy. So,                \[KE=Kd{{\cos }^{2}}\beta \] Thus,         \[K\,{{\cos }^{2}}\beta =\frac{3}{4}K\] \[\Rightarrow \]                               \[\cos \beta =\frac{\sqrt{3}}{2}\] \[\Rightarrow \]                               \[\beta ={{30}^{o}}\]