Question

A tennis ball is dropped from 1.65 m above

the ground. It rebounds to a height of 0.99 m.
With what velocity does it leave the ground?
The acceleration of gravity is 9.8 m/s².
Answer in units of m/s.

Answer 1

4.4 m/s

Step-by-step explanation:

v² = v₀² + 2a(y − y₀)

(0 m/s)² = v₀² + 2(-9.8 m/s²)(0.99 m − 0 m)

v₀ = 4.4 m/s

Answer 2

`v_i = 4.4m/s`

Step-by-step explanation:

As we know by the kinematics that final speed of the ball after it rebound will be zero as the ball will decelerate due to gravity.

here the displacement of the ball upwards is given as 0.99 m

so we will have

`v_f^2 - v_i^2 = 2 a s`

now we have

`0 - v_i^2 = 2(-9.81)(0.99)`

`v_i^2 = 19.4`

now by solving it we will have

`v_i = 4.4m/s`