Question

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

• A. ${\tan }^{-1}\left(\frac{1}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• C. ${\tan }^{-1}\left(1\right)$
• D. ${\tan }^{-1}\left(5\right)$

Answer 1

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

Horizontal velocity of bomb at ground, $V_H=500m/s$

Let Vertical velocity of bomb at ground, $V_V$ and angle of strike with horizontal is $\theta$

Using $v=u+at$ $\Rightarrow V_V=10\times 10=100m/s$

$\Rightarrow tan\theta =\frac{V_V}{V_H}=\frac{100}{500}=\frac{1}{5}$

$\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{5}\right)$