The ball was dropped from a height 20 meters
Explanation:
The given is
1. A ball is dropped from the top of a cliff
2. By the time it reaches the ground, all the energy in its gravitational
potential energy store has been transferred into its kinetic energy
store, that mean K.E = P.E
3. The ball is travelling at 20 m/s when it hits the ground
4. The gravitational field strength is 10 N/kg
We need to find the height that the ball dropped from it
The ball dropped from the top of a cliff means the initial speed is 0
→ K.E = [tex]\frac{1}{2}m(v^{2}-v_{0}^{2})[/tex]
where v is the final speed, [tex]v_{0}[/tex] in the initial speed and m
is the mass
→ v = 20 m/s and [tex]v_{0}[/tex] = 0 m/s
→ K.E = [tex]\frac{1}{2}m(20^{2}-0^{2})[/tex]
→ K.E = [tex]\frac{1}{2}m(400)[/tex]
→ K.E = 200 m joules ⇒ when the ball hits the ground
→ P.E = m g h
where g is the gravitational field strength, m is the mass and h is
the height
→ g = 10 N/kg
→ P.E = m(10)(h)
→ P.E = 10 m h joules
→ P.E = K.E
→ 10 m h = 200 m
Divide both sides by 10 m
→ h = 20 meters
The ball was dropped from a height 20 meters
Learn more
You can learn more about gravitational potential energy in brainly.com/question/1198647
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