Question

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 $(g = 10m/s^{2})$

• A. $an^{- 1}\left( \frac{1}{5} \right)$
• B. $\mathbf{\tan}\left( \frac{\mathbf{1}}{\mathbf{5}} \right)$
• C. $\tan^{- 1}(1)$
• D. $\tan^{- 1}(5)$

Answer 1

Answer: (A)

$an^{- 1}\left( \frac{1}{5} \right)$

Answer 2

Horizontal component of velocity v_x

= 500 m/s

and vertical components of velocity while striking the ground.

$v_{y} = 0 + 10 \times 10 = 100m/s$

∴ Angle with which it strikes the ground.

$\theta = \tan^{- 1}\left( \frac{v_{y}}{v_{x}} \right) = \tan^{- 1}\left( \frac{100}{500} \right) = \tan^{- 1}\left( \frac{1}{5} \right)$