If time of flight of a projectile is 10 seconds. Range is 500 meters. The maximum height attained by it will be (g=10 ms^-2)

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150 m

125 m

100 m

50 m

answer is A.

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Detailed Solution

$\begin{array}{l} T = \frac{2 usin θ}{g} = 10 \sec \Rightarrow usin θ = 50 m / s \\ H = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{(usin θ)^{2}}{2 g} \\ = \frac{50 \times 50}{2 \times 10} = 125 m \end{array}$

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