Question

When a body is projected vertically up from the ground it's potential energy to kinetic energy ratio at a point P is 1:5. If the same body is projected with half the previous velocity then at the same point P ratio of its potential energy and kinetic energy would be?

Answer 1

let the observation point P be at height 'h' from the ground.
further let the velocity of projection is 'v' .
PE at P : mgh (m= mass of the body)..............(1)

velocity of body at P : (v^2-2gh)^1/2.
then KE at P : (1/2)×m×(v^2-2gh).............(2)

Given that PE/KE=1/5.........(3)
then ,
after putting PE and KE values in (3)
we get,
v^2= 12gh...........(4)
Now if velocity of projection is decreased by half then the velocity at the point P is given by: [(v/2)^2-2gh]^1/2.
hence KE' at point P is given by:
KE' = (1/2)×m× [(v/2)^2-2gh]......(5)
putting value of v^2 from eq (4) in (5) we get:
KE'=(1/2)mgh..............(6)
As PE in both the cases is same . (obvious as P is still at height h)
Now ,
PE/KE=mgh/[(1/2)mgh]=2/1