Distance travelled in n time when a body is thrown upwards

[Govind_Balaji]

Homework Statement

Here is a question in my textbook. I saw the answer, but no explanation was there.

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

Homework 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=$\frac{v}{g}$

The Attempt at a Solution

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn2/t1.0-9/1382433_1591373084420691_3016752318298202408_n.jpg

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 $<b>v=u+at</b>$

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

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

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

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

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

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

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

 

[projjal]

man you made a slight arithmetic mistake in the last line.

 

[Govind_Balaji]

man you made a slight arithmetic mistake in the last line.

Oh yes! Thanks. I am happy at least I made only arithmetic mistake and not a concept mistake.