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A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy

Respuesta :

osikbro

Answer:

About 6.26m/s

Explanation:

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

Divide both sides by mass:

[tex]gh=\dfrac{1}{2}v^2[/tex]

Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.

[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]

Hope this helps!

snehashish65

The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Given data:

The height of vertical cliff is, h = 4.0 m.

Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,

Kinetic energy = Gravitational potential energy

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

Here,

m is the mass of rock.

v is the speed of rock.

g is the gravitational acceleration.

Solving as,

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]

Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Learn more about the conservation of energy here:

https://brainly.com/question/15707891

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