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A ball 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.)

Respuesta :

MathPhys

Answer:

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 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,

[tex]P.E_{top}=K.E_{bottom}[/tex]

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

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

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

[tex]h=\dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{20^2}{2\times10}[/tex]

[tex]h= 20 m[/tex]

Hence, the height of the cliff is 20 m.

Reference :

https://brainly.com/question/24105539

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