# 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]

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