Question

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

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

Answer 1

Answer: (C)

Let h be height of plane from the ground, then from equation of motion, we have
$h=ut+\frac{1}{2}g{t}^2$
Since, initial velocity, $u=0$
$\therefore h=\frac{1}{2}g{t}^2$
Given $t=10s,g=10m{s}^{-2}$
$\Rightarrow h=\frac{1}{2}\times 10\times (10{)}^2=500m$
Also, by equation $v^2=u^2+2gh,$
we have for $u=0,v=\sqrt{2gh}$
$\therefore v=\sqrt{2\times 10\times 500}=100m{s}^{-1}$
Hence, $\tan \theta =\frac{\text{ vertical velocity }}{\text{ horizontal velocity}}$
$=\frac{100}{500}=\frac{1}{5}$
$\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{5}\right)$