Question

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

• 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)

${{\tan }^{-1}}\left( \frac{1}{5} \right)$

Answer 2

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