Two particles are projected from the same point with the same speed $ u $ such that they have the same range $ R $, but different maximum heights $ {h_1} $ and $ {h_2} $. Which of the following is correct?
(A) $ {R^2} = 2{h_2}{h_2} $
(B) $ {R^2} = 16{h_1}{h_2} $
(C) $ {R^2} = 4{h_1}{h_2} $
(D) $ {R^2} = {h_1}{h_2} $

Hint: In this question two particles have been projected from same point with the same speed $ u $ such that, they have same range $ R $ , therefore, for the same range angle of projection will be $ \theta $ and $ 90 - \theta $, that is ,complimentary. Hence by using the concept of projectile motion we will solve this question.

Complete step by step solution:
 According to the question, particles are projected from same point with the same speed $ u $ such that, they have same range $ R $ ,
We know that $ R = \dfrac{{{u^2}\sin 2\theta }}{g} $
 $ \Rightarrow R = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g} $
Hence, for angle $ \theta $ ,
 $ {h_1} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} - - - (1) $
And For angle $ 90 - \theta $ ,
 $ {h_2} = \dfrac{{{u^2}{{\cos }^2}\theta }}{{2g}} - - - (2) $
Now, multiplying equation $ (1) $ and $ (2) $ we have,
 $ {h_1}{h_2} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \dfrac{{{u^2}{{\cos }^2}\theta }}{{2g}} $
 $ \Rightarrow {\left[ {\dfrac{{{u^2}\sin \theta \cos \theta }}{{2g}}} \right]^2} \times \dfrac{4}{4} $
 $ \Rightarrow \dfrac{1}{{16}}{\left[ {\dfrac{{{u^2}\sin 2\theta }}{g}} \right]^2} $
 $
   \Rightarrow {h_1}{h_2} = \dfrac{1}{{16}}{R^2} \\
   \Rightarrow {R^2} = 16{h_1}{h_2} \\
  $
Therefore, option (B) is the correct answer.

Note:
We must keep in mind that if two particles projected with the same range then the angle they will have between them will be complementary, that is $ \theta $ and $ 90 - \theta $ .