Question

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

Answer 1

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