Question

A stone projected up with a velocity u reaches a maximum height h. When it is at a height of
3h/4 from the ground, the ratio of KE and PE at that point is : (consider PE = 0 at the point of
projectory)
(A) 1:1
(B) 1:2
(C) 1:3
(D) 3:1​

Answer 1

Initial velocity = u

Acceleration = g

Now, at maximum height ‘h’,

02 = u2 - 2gh

=> h = u2/(2g)

If the velocity at height 3h/4 is ‘v’ then,

v2 = u2 – 2g(3h/4)

=> v2 = u2 – (3/2)gh

=> v2 = u2 – (3/2)(g)(u2)/(2g)

=> v2 = u2 – (3/4)u2

=> v2 = ¼ u2

So, KE at that height is = ½ mv2 = ½ (m)(¼ u2) = (1/8) mu2

PE at that height is = mg(3h/4) = (m)(g)(3/4)(u2)/(2g) = (3/8) mu2

So, ratio of kinetic to potential energy at that height is = (1/8) mu2 : (3/8) mu2 = 1 : 3

Answer 2

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