Question

A tennis ball is dropped on the floor from a height of 20 m . it rebound to a height of 5 m . the ball was in contact with the floor for 0.01 s .what was its average acceleration during contact?​

Answer 1

Answer :

A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m.

Time interval = 0.01s

We have to find average acceleration of ball during contact.

$\star$ Acceleration is defined as the rate of change of velocity.

It is a vector quantity having both magnitude as well as direction.

SI unit : m/s²

★ Velocity of ball just before hitting the ground :

➝ v₁ = √2gh₁

➝ v₁ = √2(10)(20)

➝ v₁ = √400

➝ v₁ = 20 m/s

★ Velocity of ball just after hitting the ground :

➝ v₂ = -√2gh₂

➝ v₂ = -√2(10)(5)

➝ v₂ = -√100

➝ v₂ = -10 m/s

Negative sign shows opposite direction.

★ Average acceleration :

⭆ a = v₁ - v₂ / ∆t

⭆ a = 20 - (-10) / 0.01

⭆ a = 30/0.01

⭆ a = 3000 m/s²