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A particle is projected from a point on the ground surface at different angles keeping the speed of projection as constant, range obtained is plotted as function of the angle of projection with the horizontal $' θ'$ . Then the maximum height attained by the particle when it is projected at an angle $60^{0}$ with the vertical : $(g = 10 m / s^{2})$

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$50 \sqrt{3} m$

125 m

62.5 m

31.25 m

answer is D.

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Detailed Solution

$250 = \frac{u^{2}}{g}$

$50 = U$

$H = \frac{u^{2} \sin^{2} 30}{2 g} = 31.25$

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