A particle is projected vertically up. It reaches a maximum height ‘H’ after time ‘T’. Its height, at any instant ‘t’ will be                                                              

Given that

A particle is projected vertically up

Initial velocity = u

Acceleration $a=-\text{g}$

We can have height ascended at any time t

Using $\text{h}=\text{ut}-\frac{1}{2}{\text{gt}}^{\text{2}}$

Or       $\text{h}=\text{vt}+\frac{1}{2}{\text{gt}}^{\text{2}}$

We know height ascended in $\text{T}=\frac{\text{u}}{\text{g}}$

is maximum height H and v = 0

We get

$\text{H}=\frac{1}{2}{\text{gT}}^{\text{2}}$

$\text{H}=\text{uT}-\frac{1}{2}{\text{gT}}^{\text{2}}$

Consider the body reached height = h

In time = t

That means in the time $=\left(\text{T}-\text{t}\right)$  sec

It would have covered remaining height

i.e. $\text{H}-\text{h}$  in the time $\left(\text{T}-\text{t}\right)$

and v = 0

Let us use $\text{h}=\text{vt}+\frac{1}{2}{\text{gt}}^2$

Then we get

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

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

We can have $\text{h}=\text{H}+\frac{1}{2}\text{g}{\left(\text{T}-\text{t}\right)}^2$

Or                    $\text{h}=\text{H}-\frac{1}{2}\text{g}{\left(\text{T}-\text{t}\right)}^2$