Question

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

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

Answer 1

Answer: (A)

Horizontal component of velocity $u_x=500m/s$ and vertical components of velocity while striking the ground.

$u_y=0+10\times 10=100m/s$



$\therefore$ Angle with which it strikes the ground,

$\theta =ta{n}^{-1}(\frac{v_y}{v_x})=ta{n}^{-1}\left(\frac{100}{500}\right)=ta{n}^{-1}(\frac{1}{5})$