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For a projectile 'R' is range and 'H' is maximum height
List - I List - II
A. R = H e. Angle of projection tan^-1(1)
B. R = 2H f. Angle of projection tan^-1(4)
C. R = 3H g. Angle of projection tan^-1(2)
D. R = 4H h. Angle of projection tan^-1(4/3)
A B C D
1. g h e f
2. h g e f
3. f g h e
4. e g f h

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answer is C.

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Detailed Solution

$θ = \tan^{- 1} (\frac{4 H}{R})$

A) $θ = \tan^{- 1} (\frac{4 H}{H}) = \tan^{- 1} (4)$

B) $θ = \tan^{- 1} (\frac{4 H}{2 H}) = \tan^{- 1} (2)$

C) $θ = \tan^{- 1} (\frac{4 H}{3 H}) = \tan^{- 1} (\frac{4}{3})$

D) $θ = \tan^{- 1} (\frac{4 H}{4 H}) = \tan^{- 1} (1)$

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