Two stones are thrown vertically upwards simultaneously with their initial velocities ${{u}_{1}}$ and ${{u}_{2}}$ respectively. Prove that the heights reached by them would be in the ratio of
A. $u_{1}^{2}:u_{2}^{2}$
B. ${{u}_{1}}:{{u}_{2}}$
C. $u_{2}^{2}:u_{2}^{2}$
D. ${{u}_{2}}:{{u}_{1}}$

Hint: Assume upward acceleration as –g and downward acceleration is +g. use a kinematic equation to solve this question. Consider the value of final velocity as zero for both the stones and Initial velocity must be different.

Complete Step-by-Step solution:
Find the ratio of height i.e. the height gained by stones
Stone 1:
When it reaches to maximum level then final velocity will be zero,
V=0 and u=${{u}_{1}}$
Now find out height.
We already have assumed that upward acceleration as –g.
i.e. a= -g
Now use the kinematic equation. Third equation of motion,
${{V}^{2}}-{{u}^{2}}=2\times a\times h$
Put values of V and u
$0-u_{1}^{2}=2\times (-g)\times {{h}_{1}}$
${{h}_{1}}=\dfrac{u_{1}^{2}}{2g}$
Similarly,
For stone 2:
When it reaches to maximum level then final velocity will be zero,
V=0 and u=${{u}_{2}}$
Now find out height.
We already have assumed that upward acceleration as –g.
i.e. a= -g
Put values of V and u in kinematic equation,
$\begin{align}
  & 0-u_{2}^{2}=2\times (-g)\times {{h}_{2}} \\
 & {{h}_{2}}=\dfrac{u_{2}^{2}}{2g} \\
\end{align}$
Take ratio of both of stone’s height
$\begin{align}
  & \dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{{{h}_{1}}=\dfrac{u_{1}^{2}}{2g}}{{{h}_{2}}=\dfrac{u_{2}^{2}}{2g}} \\
 & \dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{u_{1}^{2}}{u_{2}^{2}} \\
\end{align}$
So the answer is option A.

Note: Always use a kinematic equation to solve this kind of question where the question has given initial and final velocity or acceleration. When an object throws up then gravitational force is the one which will always act no matter what kind of body or object it is.