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. $ta{n}^{-1}\left(\frac{1}{5}\right)$
• B. $tan\left(\frac{1}{5}\right)$
• C. $ta{n}^{-1}\left(1\right)$
• D. $ta{n}^{-1}\left(5\right)$

Answer 1

The correct option is A $ta{n}^{-1}\left(\frac{1}{5}\right)$

Horizontal component of velocity $v_x$= 500 m/s and vertical components of velocity while striking the ground.
$v_y=0+10\times 10=100m/s$
$\therefore$ Angle with which it strikes the ground.
$\theta =ta{n}^{-1}\left(\frac{v_y}{v_x}\right)=ta{n}^{-1}\left(\frac{100}{500}\right)=ta{n}^{-1}\left(\frac{1}{5}\right)$