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CGP EDU Academic Team
Published on: September 12, 2026
A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which it strikes the ground will be \(\left(g = 10 \, m/s^{2}\right)\)
Text Solution
Verified by ExpertsThe correct answer is:
A
Horizontal component of velocity v x = 500 m / s
and vertical components of velocity while striking the ground.
\(v_y = 0 + 10 \times 10 = 100\,m/s\)
∴ ∴ Angle with which it strikes the ground.
\(\theta = \tan^{-1} \left( \frac{v_y}{v_x} \right) = \tan^{-1} \left( \frac{100}{500} \right) = \tan^{-1} \left( \frac{1}{5} \right)\)
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