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.