Question

a stone projected vertically upward with velocity U reaches a maximum height H.when it is at a height of 3 h / 4 from the ground the ratio of PE and KE at the point is ?

Answer 1

Answer:

Explanation:

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

Answer:

→ 1:3

Explanation:

→ 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