A body is projected up with a speed ‘u’ and the time taken by it is ‘T’ to reach the maximum height H. Pick out the correct statement

Given that

A body is projected vertically up initial velocity$=\text{u}$  

Acceleration $a=-\text{g}$  

It will ascend up until its final velocity becomes zero i.e.$\text{v}=\text{o}$ then $\text{t}=?$  $\text{h}=?$  

We get $\text{v}=\text{u}-\text{gt}$

$0=\text{u}-\text{gt}$

$\text{T}=\text{t}=\frac{\text{u}}{\text{g}}$

So it will ascend for $\text{t}=\frac{\text{u}}{\text{g}}$

Maximum height ascended

$\text{h}=\text{ut}-\frac{1}{2}{\text{gt}}^{\text{2}}\because \text{t}=\frac{\text{u}}{\text{g}}$

$\text{h}=\frac{{\text{u}}^{\text{2}}}{\text{g}}-\frac{1}{2}\text{g}\left(\frac{{\text{u}}^{\text{2}}}{{\text{g}}^{\text{2}}}\right)$

${\text{h}}_{\text{max}}=\text{H}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}$

We get

$\left(i\right)\text{T}=\text{u/g}\text{H}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}=\frac{1}{2}{\text{gt}}^{\text{2}}\text{v}=0,\text{v}=\text{u}-\text{gT}$

$\left(\text{ii}\right)\text{t}=\text{T/2}\text{H}=\frac{{\text{u}}^{\text{2}}}{\text{8g}}=\frac{\text{H}}{\text{4}}\text{v}=\text{u/2}$

$\left(iii\right)\text{t}=\frac{\text{T}}{\sqrt{\text{2}}}\text{h}=\frac{{\text{u}}^{\text{2}}}{\text{4g}}=\frac{\text{H}}{\text{2}}\text{v}=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\text{u}$

$\left(iv\right)\text{t}=2\text{T}\text{h}=\frac{{\text{u}}^{\text{2}}}{\text{g}}=\text{2H}\text{v}=-\text{u}$

We get only option (B) is true