Q. 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) $\tan^{- 1}\left( \frac{1}{2} \right)$
C) $\tan^{- 1}(1)$
D) $\tan^{- 1}(5)$
Explanation
(a) Horizontal component of velocity $\upsilon_{x} = 500\ m/s$ and vertical components of velocity while striking the ground
$\upsilon_{y} = 0 + 10 \times 10 = 100\ m/s$
$\therefore$ Angle with which it strikes the ground
$\theta = \tan^{- 1}\left( \frac{\upsilon_{y}}{\upsilon_{x}} \right) = \tan^{- 1}\left( \frac{100}{500} \right) = \tan^{- 1}\left( \frac{1}{5} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.