Question

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

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

Answer 1

Answer: (A)

Horizontal component of velocity $u_{x} = 500\, m/s$ and vertical components of velocity while striking the ground.
$u_{y} = 0 + 10 \times 10 = 100 \,m/s$

$\therefore $ Angle with which it strikes the ground,
$\theta=tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}\left(\frac{100}{500}\right)=tan^{-1}(\frac{1}{5})$