Question

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

Answer 1

`PE:KE= 1:1.25`

Explanation:

`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`