# When a body is projected vertically up from the ground, its potential energy and kinetic energy at a point P are in the ratio 1:5. if the same body is projected up with half the previous velocity then at the same point P the ratio will be what?

## emphasized text

Nov 5, 2015

$P E : K E = 1 : 1.25$

#### Explanation:

$K E = \frac{1}{2} m {v}^{2}$

$P E = m g h$

@$P$:

$\frac{\frac{1}{2} m {v}^{2}}{m g h} = 5$

$\therefore \frac{{v}^{2}}{2 g h} = 5 \text{ } \textcolor{red}{\left(1\right)}$

The velocity is now halved but $h$ remains the same so we can write:

$\frac{{\left(\frac{v}{2}\right)}^{2}}{2 g h} = R$

Where $R$ is the ratio of $\frac{K E}{P E}$

$\therefore \frac{\frac{{v}^{2}}{4}}{2 g h} = R$

$\therefore \frac{{v}^{2}}{8 g h} = R \text{ } \textcolor{red}{\left(2\right)}$

Now we can divide $\textcolor{red}{\left(1\right)}$ by $\textcolor{red}{\left(2\right)} \Rightarrow$

$\frac{{v}^{2}}{2 g h} . \frac{8 g h}{{v}^{2}} = \frac{5}{R}$

Which cancels down to:

$4 = \frac{5}{R}$

$R = \frac{5}{4}$

$R = 1.25 = \frac{K E}{P E}$

So the ratio $P E : K E = 1 : 1.25$