Question

A ball is dropped from a height of 50 m then at what height the kinetic
energy and potential energies are equal.​

Answer 1

Answer:

Let h be the height from which the ball has been dropped

So, loss in kinetic energy ΔK=50%ofmgh=0.5mgh

Let H is the height from where the ball is placed.

U=mgH

As the ball is released the potential energy is converted into kinetic energy.

When the ball strikes the earth, net potential energy is converted into kinetic energy as

K=mgH

So, loss in kinetic energy is

ΔK=

100

50

mgH

ΔK=

2

mgH

also,

ΔK=K

f

−K

i

K

f

=ΔK−K

i

K

f

=

2

mgH

−mgH

K

f

=

2

mgH

This kinetic energy will take the ball to the height where the potential energy of the ball will be

U

′

=

2

mgH

Hence, the ball will rise to half the height from where it was dropped.

Hence, the height attained =

2

1

Answer 2

Answer:

25 m

Explanation:

Potential energy is the energy a body has because of it's position. For example, a spring which is stretched or compressed will have some energy, due to which it will try to regain its former shape. A body placed at some height 'h' will want to come back to the ground because it will possess gravitational potential energy.

Kinetic energy is the energy due to it's speed. A fast moving object will have more energy than a slow moving object of same mass.

Gravitational Potential Energy: mgh

Kinetic energy: $mv^2/2$

So, we want to find the moment when both K.E and P.E are equal.

So, K.E = P.E

mgh = $mv^2/2$

h is some height from the ground, which we have to find.

Now, you have to remember that total energy of the ball is ALWAYS CONSERVED.

So, mgh + $mv^2/2$ = Initial total energy

Initially, total energy was only potential energy.

Initial total energy = mg(50) = 50mg

50mg = mgh + $mv^2/2$

But, since they gave that P.E and K.E are equal:

50mg = 2mgh

h = 25 m

Therefore at 25 m from the ground, both kinetic energy and potential energy are equal.

Hope it helps!