Answer:
The ball hits the ground with velocity 6.21 m/s downward
Explanation:
A tennis ball is dropped from 1.97 m above the ground
It rebounds to a height of 0.945 m
We need to find its velocity when it hits the ground
The acceleration of gravity is 9.8 m/s
The given is:
→ Initial velocity = 0 m/s
→ Height = 1.97 meters
→ Acceleration = 9.8 m/s²
→ v = ?
We need a rule satisfies the given
→ v² = u² + 2 a h
where u is the initial velocity, v is the final velocity, a is the acceleration
and h is the height
Substitute the values above in the rule
→ v² = (0)² + 2(9.8)(1.97)
→ v² = 38.612
Take √ for both sides
→ v = ± [tex]\sqrt{38.612}[/tex]
→ v = - 6.21
We put negative sign for the value of velocity because it's downward
The ball hits the ground with velocity 6.21 m/s downward
