1. The problem statement, all variables and given/known data
Here is a question in my text book. I saw the answer, but no explanation was there.

A stone is thrown vertically upward with an initial velocity $v_0$. The distance travelled by it in time $\frac{1.5v_0}{g}$ is ______________.
Answer is $\frac{5v_0^2}{8g}$.

2. Relevant equations
Max. height reached by a body thrown upwards with an initial velocity v is $\frac{v^2}{2g}$
Time taken to reach the max.height with initial velocity v=$\frac{v}{g}$

3. The attempt at a solution

Time taken to travel AB+BC=$\frac{1.5v_0}{g}$

We know that time taken to travel AB=$\frac{v_0}{g}$

∴Time taken to travel BC=$\frac{1.5v_0}{g} – \frac{v_0}{g}$
===================$\frac{v_0}{2g}$

Initial velocity at B in BC=u=0

Acceleration=a=g

We know $v=u+at$

$v=0+g.\frac{v_0}{2g}=\frac{v_0}{2}$

$v^2=u^2+2as$

$\frac{v_0^2}{2^2}=0^2+2.g.s$

$\frac{v_0^2}{4}=2gs$

$s=\frac{v_0^2}{8g}$

Distance of AB+BC=$\frac{v_0^2}{2g}+\frac{v_0^2}{8g}$

==============$\frac{4v_0^2}{8g}$

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