A stone is thrown upward with a speed ‘u’ from the top of the tower reaches the ground with velocity of 3 u. The height of the tower is 

For Projected body

$\text{u}=\text{u,v}=\text{0,a=}-g,\text{s}={\text{h}}_{\text{1}}$

${\text{h}}_{\text{1}}=\frac{{\text{v}}^{\text{2}}-{\text{u}}^2}{\text{2g}}=\frac{{\text{0}}^{\text{2}}-{\text{u}}^2}{\text{2g}}=\frac{-{\text{u}}^2}{(-\text{2g})}=\frac{{\text{u}}^2}{\text{2g}}$

For Falling body

$\text{u}=0\text{,v}=3\text{u,a}=g,\text{s}={\text{h}}_{\text{2}}$

${\text{h}}_{\text{2}}=\frac{{\left(3u\right)}^{\text{2}}-{\left(0\right)}^2}{\text{2g}}=\frac{{\text{9u}}^2}{\text{2g}}$

$\therefore \text{h}={\text{h}}_1-{\text{h}}_{\text{2}}$

${\text{h}}_{\text{2}}=\frac{{\text{9u}}^2}{\text{2g}}-\frac{{\text{u}}^2}{\text{2g}}=\frac{{\text{8u}}^2}{\text{2g}}=\frac{{\text{4u}}^2}{\text{g}}$

Second method:

$\text{u\hspace{0.17em}=\hspace{0.17em}u,\hspace{0.17em}\hspace{0.17em}v\hspace{0.17em}=\hspace{0.17em}-3u}$

$v^2-u^2=2as$

$\Rightarrow {\text{(-3u)}}^{\text{2}}{\text{-u}}^{\text{2}}\text{=2gh}$

$\Rightarrow 8u^2=2gh$

$\Rightarrow \frac{{\text{8u}}^2}{\text{2g}}=\frac{{\text{4u}}^2}{\text{g}}$