91Ó°ÊÓ

The slower time taken is 9 s.

Let elapsed time to peak point for faster stone be \({t_1}\)

The velocity of the faster stone

\({t_1} = \frac{{3v}}{g}\)

Let elapsed time to peak point for slower stone be

The velocity of the slower stone

\({t_2} = \frac{v}{g}\)

Let's find the ratio

\(\begin{array}{l}\frac{{{t_1}}}{{{t_2}}} = \frac{{3v}}{g} \times \frac{g}{v}\\\frac{{{t_1}}}{{{t_2}}} = 3\end{array}\)

Since the movement is symmetric, the time to reach

The maximum point is equal to half of the duration of the flight.

\(\begin{array}{c}{t_1} = \frac{{10}}{2}\\ = 5\,{\rm{s}}\\\frac{5}{{{t_2}}} = 3\\{t_2} = \frac{5}{3}\,{\rm{s}}\end{array}\)

Flight time for slower stone is,

\(\begin{array}{l} = 2 \times \frac{5}{3}\\ = 3.33\,{\rm{s}}\end{array}\)

Lets slower stone's maximum height

\(H = \frac{{{v^2}}}{{2g}}\)

Faster stone's maximum height

\(x = \frac{{3{v^2}}}{{2g}}\)

Now equating we get

\(\begin{array}{l}\frac{H}{x} = \frac{{{v^2}}}{{2g}} \times \frac{{2g}}{{9{x^2}}}\\x = 9H\end{array}\)