Q. A bomber plane moves horizontally with a speed of $500ms^{- 1}$ and a bomb released from it, strikes the ground in $10\ s$. Angel at which it strikes the ground will be $(g = 10ms^{- 2})$
A) $tan^{- 1}\left( \frac{1}{5} \right)$
B) $tan^{\ }\left( \frac{1}{5} \right)$
C) $\tan^{- 1}{(1)}$
D) $\tan^{- 1}{(5)}$
Explanation
(a) Horizontal component of velocity $v_{x} = 500ms^{- 1}$ and vertical components of velocity while striking the ground
$v_{y} = 0 + 10 \times 10 = 100\ ms^{- 1}$
$\therefore\ $ Angle with which it strikes the ground
$\theta = \tan^{- 1}{\left( \frac{v_{y}}{v_{x}} \right) = \tan^{- 1}\left( \frac{100}{500} \right)}$
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$\theta = \tan^{- 1}\left( \frac{1}{5} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.