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Ostrovityanka [42]

2 years ago

14

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

gy store has been transferred into its kinetic energy store. If the ball is travelling at 20 m/s when it hits the ground, what height was it dropped from? (Assume that the gravitational field strength is 10 N/kg.)

1 answer:

Bingel [31]2 years ago

8 0

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 = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final speed, $v_{0}$ in the initial speed and m

is the mass

→ v = 20 m/s and $v_{0}$ = 0 m/s

→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ 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

#LearnwithBrainly

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