Question:medium
A particle is projected at an angle of \( 30^\circ \) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \( h_0 \) and height traversed in the last second, before it reaches the maximum height, is \( h_1 \). The ratio \( \frac{h_0}{h_1} \) is __________. [Take \( g = 10 \, \text{m/s}^2 \)]
A particle is projected at an angle of \( 30^\circ \) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \( h_0 \) and height traversed in the last second, before it reaches the maximum height, is \( h_1 \). The ratio \( \frac{h_0}{h_1} \) is __________. [Take \( g = 10 \, \text{m/s}^2 \)]
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For projectile motion, use the vertical component of velocity to calculate heights in different time intervals.
Updated On: Jan 14, 2026
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Correct Answer: 5
Solution and Explanation
The vertical velocity component is calculated as: \[60 \sin 30^\circ = 30 \, \text{m/s}\]The distance covered in the first second, \( S_1 \), is: \[S_1 = 30 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 25 \, \text{m}\]The distance covered in the final second, \( S_3 \), is: \[S_3 = 30 + \left( \frac{-10}{2} \right) \times (2 \times 3 - 1) = 5 \, \text{m}\]The resulting ratio is: \[\frac{S_1}{S_3} = \frac{25}{5} = 5\]
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