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

Question

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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.)

Physics

1 answer:

ValentinkaMS [3.4K] · Answer 1 · 2026-03-01T04:31:04+03:00

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final velocity, $v_{0}$ is the initial velocity, and m is the mass.

<p→ v = 20 m/s and $v_{0}$ = 0 m/s.<p→ 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 strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

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