question_answer

A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which it strikes the ground will be \[(g=10\,\,m/{{s}^{2}})\]

A) \[{{\tan }^{-1}}\left( \frac{1}{5} \right)\]

B) \[\tan \,\left( \frac{1}{5} \right)\]

C) \[{{\tan }^{-1}}(1)\]

D) \[{{\tan }^{-1}}(5)\]

Correct Answer: A

Solution :

[a]        Horizontal component of velocity vx= 500 m/s and vertical components of velocity while striking the ground. \[{{v}_{y}}=0+10\times 10=100\,m/s\]      \[\therefore \] Angle with which it strikes the ground. \[\theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)={{\tan }^{-1}}\left( \frac{100}{500} \right)={{\tan }^{-1}}\left( \frac{1}{5} \right)\]