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?

Question

emphasized text

13436 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-05T23:26:55

$PE:KE= 1:1.25$

$KE=1/2mv^2$

$PE=mgh$

@ $P$ :

$(1/2mv^2)/(mgh)=5$

$:.(v^2)/(2gh)=5 " "color(red)((1))$

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

$((v/2)^2)/(2gh)=R$

Where $R$ is the ratio of $(KE)/(PE)$

$:.((v^2)/(4))/(2gh)=R$

$:.(v^2)/(8gh)=R " "color(red)((2))$

Now we can divide $color(red)((1))$ by $color(red)((2))rArr$

$(v^2)/(2gh).(8gh)/(v^2)=5/R$

Which cancels down to:

$4=5/R$

$R=5/4$

$R=1.25=(KE)/(PE)$

So the ratio $PE:KE=1:1.25$