Question

A stone is projected vertically up to reach a maximum height h.The ratio of its kinetic energy to potential energy at a height 4h/5 will be

Answer 1

Gravitational potential energy near the surface of earth( where acceleration due to gravity has constant value) is directly proportional to height from the surface. At a height h potential energy = mgh.
Kinetic energy = 0
At a height 4h/5 potential energy = 4/5mgh.
The difference mgh - 4/5mgh= 1/5mgh is the kinetic energy.
Ratio of kinetic energy to potential energy is 1:4.

Answer 2

Answer:

The ratio of kinetic energy to potential energy at height ⅘h above ground is 1 : 4 .

Explanation:

• Let, the mass of the stone be m and the point from where the stone is projected vertically upward just above the ground be point A.
• Let, the maximum height attend by the stone be h at point B.
• The potential energy of any particle at height h from the surface of ground is given by mgh.
• The kinetic energy of any particle moving with velocity v is given by ½mv² .
• Now, let the velocity of a stone at point A be v. The potential energy at point A is zero and the kinetic energy at point A is given by ½mv² .
• This kinetic energy changes into potential energy as the stone moves vertically upward.
• At point B, the kinetic energy is zero and the potential energy is mgh.
• Let, the point ⅘h above the surface be point C. The velocity of the stone at this point be v1.
• Potential energy at point C , U(c) is given by ⅘mgh.
• Kinetic energy of stone at point C is given by

$K(c) = \frac{1}{2} m( {v1}^{2} ) - - - (1)$

From the third equation of motion we can get value of v1

Initial velocity, u = 0

Final velocity= v1

Distance between point B and C = h - ⅘h = ⅕h

${v1}^{2} = 2g( \frac{1}{5} h) \\ = \frac{2}{5} gh - - - (2)$

Putting value of equation (2) in equation (1), we get

$K(c) = \frac{1}{2} m( \frac{2}{5}gh) \\ = \frac{1}{5} mgh$

Ratio of kinetic energy to potential energy is given by :

$\frac{K(c)}{U(c)} = \frac{ \frac{1}{5} mgh}{ \frac{4}{5} mgh} \\ = \frac{1}{4} \\ = 1:4$

Hence, the ratio of kinetic energy to potential energy at height ⅘h above ground is 1 : 4 .