Aball is dropped from the top of a cliff. by the time it reaches the ground, all of its gravitational potential energy has been transferred into kinetic energy. if the ball is travelling at 20m/s when it hits the ground, what height was it dropped from? (assume that the gravitational field strength is 10n/kg.)

The height of the cliff is 20 m.

Explanation:

Given that,

Velocity of the ball v = 20 m/s

Gravitational field strength g= 10 N/Kg

It reaches the ground, all of its gravitational potential energy has been transferred into kinetic energy.

According to the conservation of energy,

Where,

m = mass of the ball

g = gravitational field strength

v= velocity of the ball

h = height

We need to calculate the height of the cliff

Using conservation of energy

Put the value into the formula

Hence, the height of the cliff is 20 m.

Reference :

link

20 m

Explanation:

Initial potential energy = final kinetic energy

mgh = 1/2 mv²

gh = 1/2 v²

h = v² / (2g)

Given v = 20 m/s and g = 10 m/s²:

h = (20 m/s)² / (2 × 10 m/s²)

h = 20 m

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.