Question

A bomber plane moves horizontally with a speed of $600m/s$ and a bomb released from it, strikes the ground in $10s$. The angle with horizontal at which it strikes the ground will be

• A. ${\tan }^{-1}\left(1/2\right)$
• B. ${\tan }^{-1}\left(1/6\right)$
• C. ${\tan }^{-1}\left(4/5\right)$
• D. ${\tan }^{-1}\left(3/4\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(1/6\right)$

Bomb was released so initial velocity will be $zero$ so the equation for $vertical$ motion will be $v=gt$

putting $t=10sec$ we get the $vertical$ velocity at the moment of hit as $v=10\times 10=100m/s$

At the $moment$ of the release the bomb will $acquire$ the $horizontal$ velocity of the plane which will remain $constant$

as $u=600m/s$

angle with horizontal $\theta =ta{n}^{-1}\frac{V_{vertical}}{V_{horizontal}}=ta{n}^{-1}\frac{v}{u}=ta{n}^{-1}\frac{100}{600}$

or $\theta =ta{n}^{-1}\frac{1}{6}$