Question

A bomber plane moves horizontally with a speed of $500\text{ ms}^{-1}$ and a bomb released from it strikes the ground in $10\text{ sec}$. Angle with which it strikes the ground will be ($g = 10\text{ ms}^{-2}$):

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

Hint:

For horizontal projection, the angle of impact with the horizontal is always $\theta = \tan^{-1}\left(\frac{gt}{u}\right)$.
This simple formula avoids the need to resolve intermediate displacement vectors.

Answer 1

The Correct Option is A

Step 1: Write the trajectory in parametric form.
Taking the release point as origin, the horizontal distance is $x = vt$ and the vertical fall is $y = \frac{1}{2}gt^2$, since the initial vertical speed is zero.
Step 2: Find the slope of the path.
The angle the bomb makes with the ground at impact is the slope of its path there, $\tan\theta = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{gt}{v}$.
Step 3: Substitute the numbers.
With $v = 500\text{ ms}^{-1}$, $g = 10\text{ ms}^{-2}$ and $t = 10\text{ s}$, \[ \tan\theta = \frac{10\times10}{500} = \frac{100}{500} = \frac{1}{5} \]
\[ \boxed{\theta = \tan^{-1}\left(\frac{1}{5}\right)} \]