Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A stone is thrown from point O on the ground with velocity
. Its angular
velocity about O, when the stone is at maximum height of its trajectory, is
Text Solution
Verified by ExpertsThe correct answer is:
C
Step 1: Determine the horizontal () and vertical components of the initial velocity. Here, = 5 m/s (horizontal), and vertical component at max height = 0.
Step 2: Angular velocity at maximum height can be calculated using the formula: \( \omega = \frac{v}{r} \)
At maximum height, the radius (distance from O to the stone) can be found using: \( r = \frac{v^2}{g} = \frac{10^2}{9.81} \)
Calculating gives: \( r \approx 10.2 \, m \)
Step 3: Substitute into the angular velocity formula. The total velocity at maximum height is only horizontal, thus: \( v = 5 \, m/s \)
Therefore: \( \omega = \frac{5}{10.2} \approx 0.49 \, rad/s \approx \frac{1}{\sqrt{2}} \, rad/s \)
Hence, Option C is correct.
Step 2: Angular velocity at maximum height can be calculated using the formula: \( \omega = \frac{v}{r} \)
At maximum height, the radius (distance from O to the stone) can be found using: \( r = \frac{v^2}{g} = \frac{10^2}{9.81} \)
Calculating gives: \( r \approx 10.2 \, m \)
Step 3: Substitute into the angular velocity formula. The total velocity at maximum height is only horizontal, thus: \( v = 5 \, m/s \)
Therefore: \( \omega = \frac{5}{10.2} \approx 0.49 \, rad/s \approx \frac{1}{\sqrt{2}} \, rad/s \)
Hence, Option C is correct.
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