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Two bodies are thrown from the same point with the same velocity of $50 {ms}^{- 1}$ . If their angles of projection are complimentary angles and the difference of maximum heights is 30m, their maximum heights $(g = 10 {ms}^{- 2})$

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30m and 60m

25m and 55m

47.5m and 77.5m

50m and 80m

answer is B.

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Detailed Solution

H_{1} + H_{2} = \frac{u^{2}}{2 g}

$= \frac{50^{2}}{2 \times 10} = \frac{2500}{20} = 125$

$H_{1} - H_{2} = 30$

$H_{1} + H_{2} = 125$

$\underline{\begin{matrix} H_{1} - H_{2} = 30 \\ - - - \end{matrix}}$

$2 H_{2} = 95$

$H_{2} = \frac{95}{2} = 47.5$

$H_{1} + H_{2} = 125$

$H_{1} + 47.5 = 125$

$H_{1} = 125 - 47.5$

$H_{1} = 77.5 m$

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