Question

A ball is dropped from a height of 80 metres and after striking the ground it velocity is became half what is its height after striking

Answer 1

Given :-

▪ A ball is dropped from a height of 80 m and after striking the ground its velocity became half.

To Find :-

▪ Height at which the ball rebounded after striking.

Solution :-

According to the conservation of energy, All the potential energy stored at the height of 80 m will be converted into kinetic energy at the time of striking.

So,

⇒ mass × g × h = 1/2 × mass × velocity²

⇒ 2 × g × h = v²

⇒ v = √(2gh)

Let's find the velocity of the ball at the time of striking,

⇒ v = √( 2 × 10 × 80) [ g = 10 m/s² ]

⇒ v = √(1600)

⇒ v = 40 m/s

Given that after striking the ball, the velocity of the ball becomes half,

Similarly, The kinetic energy at that time will be converted into Potential energy at height h, according to the conservation of energy.

So,

⇒ m × g × h = 1/2 × m × (v/2)²

⇒ 2 × 10 × h = (20)²

⇒ 20h = 400

⇒ h = 20 m

Hence, The ball will go to a height of 20 meters after striking.

Answer 2

Answer :

Let ball is dropped from a height of H m$.$

We know that, Potential energy is taken zero at reference point or at ground.

Initial velocity of ball = zero.

★ This question can be solved by using concept of mechanical energy conservation.

Potential energy of body of mass m at height H from ground is given by : U = mgH

Applying concept of mechanical energy conservation,

➝ U₁ + K₁ = U₂ + K₂

➝ mgH + 0 = 0 + 1/2 mv²

➝ 1/2mv² = m × 10 × 80

➝ 1/2 v² = 800

➝ v² = √1600

➝ v = 40 m/s

Let ball is raised by height h after striking to the ground.

After striking to the ground velocity of ball becomes half of its initial value.

• v' = 40/2 = 20 m/s

Again applying concept of mechanical energy conservation,

➝ U₂ + K₂ = U₃ + K₃

➝ 0 + 1/2 mv'² = mgh + 0

➝ 1/2v'² = gh

➝ 1/2(20)² = 10h

➝ 200 = 10h

➝ h = 20m

∴ Ball will attain 20m of height after striking to the ground.