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 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 ($g = 10\text{ ms}^{-2}$):
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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.
This simple formula avoids the need to resolve intermediate displacement vectors.
Updated On: Jul 22, 2026
- $\tan^{-1}\left(\frac{1}{5}\right)$
- $\tan^{-1}\left(\frac{1}{2}\right)$
- $\tan^{-1}(2)$
- $\tan^{-1}(5)$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
A bomb is released from a plane flying horizontally.
We need to find the angle at which the bomb's velocity vector strikes the ground, given the plane's horizontal speed and the fall time.
Step 2: Key Formula and Approach:
The motion of the bomb is a horizontal projectile motion.
The horizontal component of velocity ($v_x$) remains constant.
The vertical component of velocity ($v_y$) increases due to gravity:
\[ v_y = u_y + gt = gt \] The angle $\theta$ with the horizontal is given by:
\[ \tan\theta = \frac{v_y}{v_x} \]
Step 3: Detailed Explanation:
• Horizontal component of velocity ($v_x$):
\[ v_x = 500\text{ ms}^{-1} \]
• Vertical component of velocity ($v_y$):
Since the initial vertical velocity is zero ($u_y = 0$):
\[ v_y = g \times t = 10\text{ ms}^{-2} \times 10\text{ s} = 100\text{ ms}^{-1} \]
• Calculate the striking angle $\theta$:
\[ \tan\theta = \frac{v_y}{v_x} = \frac{100}{500} = \frac{1}{5} \] \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]
Step 4: Final Answer:
The angle at which the bomb strikes the ground is $\tan^{-1}\left(\frac{1}{5}\right)$, which corresponds to Option (A).
A bomb is released from a plane flying horizontally.
We need to find the angle at which the bomb's velocity vector strikes the ground, given the plane's horizontal speed and the fall time.
Step 2: Key Formula and Approach:
The motion of the bomb is a horizontal projectile motion.
The horizontal component of velocity ($v_x$) remains constant.
The vertical component of velocity ($v_y$) increases due to gravity:
\[ v_y = u_y + gt = gt \] The angle $\theta$ with the horizontal is given by:
\[ \tan\theta = \frac{v_y}{v_x} \]
Step 3: Detailed Explanation:
• Horizontal component of velocity ($v_x$):
\[ v_x = 500\text{ ms}^{-1} \]
• Vertical component of velocity ($v_y$):
Since the initial vertical velocity is zero ($u_y = 0$):
\[ v_y = g \times t = 10\text{ ms}^{-2} \times 10\text{ s} = 100\text{ ms}^{-1} \]
• Calculate the striking angle $\theta$:
\[ \tan\theta = \frac{v_y}{v_x} = \frac{100}{500} = \frac{1}{5} \] \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]
Step 4: Final Answer:
The angle at which the bomb strikes the ground is $\tan^{-1}\left(\frac{1}{5}\right)$, which corresponds to Option (A).
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