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?

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1 Answer
Nov 5, 2015

Answer:

#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#