Question

A ball at rest is dropped from a height of 12m.it leaves 25% of it's kinetic energy on strucking the ground.find the height to which it bounces

Answer 1

Given that, a ball at test is dropped from a height of 12 m.

{ height of ball i.e. h is 12 m and acceleration due to gravity i.e. g is 10 m/s² }

Potential energy = mgh

= m × 10 × 12

= 120 × m

By Conversation of Energy,

→ Potential Energy = Kinetic Energy

120 × m = 1/2 mv²

Also given that, the ball losses 25% of its kinetic energy.

= 25/100 × 120 × m

= 30 × m (Lost in energy)

Now,

Remaining energy = Potential energy - Lost in energy

= (120 × m) - (30 × m)

= m(120 - 30)

= 90 × m

We have to find the height to which the ball bounces.

From the above calculations, the ball possesses the energy of (90 × m) which is remaining energy and with this energy, the ball bounces to height 'h'.

Now,

Potential energy = mgh

Substitute the known values

→ 90 × m = m × 10 × h

→ 90 = 10h

→ h = 9

Therefore, 9 m is the height of the ball.

Answer 2

$\rule{300}{2}$

GIVEN :-

≛ A ball at rest is dropped from a height of 12 m.

≛ It leaves 25% of it's kinetic energy on striking the ground.

• Acceleration due to gravity(g) = 10 m/s²

• Height(h) = 12 m

TO FIND :-

The height to which it bounces.

FORMULAS TO BE USED :-

Potential energy = mgh

Kinetic Energy = 1/2 mv²

SOLUTION :-

Applying the formula :-

Potential energy = mgh = m × 10 × 12

⇒P.E. = 120m

By conservation of energy,

P.E. = K.E.

⇒120m = 1/2 mv²

(Initial potential energy = final kinetic energy)

A/q,

Ball loses 25% of (final K.E.)

⇒25% of 120m is lost.

⇒25/100 × 120m is lost

⇒30m is lost

Now, the energy with which the ball bounces back =

final kinetic energy - Lost in energy

= 120m - 30m

= 90m

Now, the ball bounces to height 'h'.

Or we can say, the ball obtains a potential energy of 90m.

⇒P.E. = mgh

⇒90m = m × 10 × h

⇒90 = 10h

⇒h = 9 meters

➠ So, the height to which the ball bounces back is 9 m.

$\rule{300}{2}$