Solution:
Using third equation of motion: \(v^2 = u^2 - 2g(H/2) \Rightarrow 10^2 = u^2 - gH\). At max height, \(0 = u^2 - 2gH \Rightarrow u^2 = 2gH\). Thus, \(100 = 2gH - gH = gH \Rightarrow H = 10\text{ m}\).
A particle is thrown vertically upward from ground. Its velocity at half of the height is \(10\text{ m/s}\), then maximum height attained by it (\(g = 10\text{ m/s}^2\))
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