Distance travelled in n time when a body is thrown upwards

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 [itex]v_0[/itex]. The distance travelled by it in time [itex]\frac{1.5v_0}{g}[/itex] is ______________.

Answer is [itex]\frac{5v_0^2}{8g}[/itex].





2. Relevant equations

Max. height reached by a body thrown upwards with an initial velocity v is [itex]\frac{v^2}{2g}[/itex]

Time taken to reach the max.height with initial velocity v=[itex]\frac{v}{g}[/itex]





3. The attempt at a solution





Time taken to travel AB+BC=[itex]\frac{1.5v_0}{g}[/itex]



We know that time taken to travel AB=[itex]\frac{v_0}{g}[/itex]



∴Time taken to travel BC=[itex]\frac{1.5v_0}{g} - \frac{v_0}{g}[/itex]

===================[itex]\frac{v_0}{2g}[/itex]





Initial velocity at B in BC=u=0



Acceleration=a=g



We know [itex]v=u+at[/itex]



[itex]v=0+g.\frac{v_0}{2g}=\frac{v_0}{2}[/itex]



[itex]v^2=u^2+2as[/itex]



[itex]\frac{v_0^2}{2^2}=0^2+2.g.s[/itex]



[itex]\frac{v_0^2}{4}=2gs[/itex]



[itex]s=\frac{v_0^2}{8g}[/itex]



Distance of AB+BC=[itex]\frac{v_0^2}{2g}+\frac{v_0^2}{8g}[/itex]



==============[itex]\frac{4v_0^2}{8g}[/itex]