a stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes the ground is v

Answer :

Answer:

[tex]v=\sqrt{2gh}[/tex]

Explanation:

We could use conversation of energy. Total distance the stone will cover will be

[tex]h=\frac{1}{2} g t^{2}[/tex]

the Final velocity will be

[tex]v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}[/tex]

The velocity will be [tex]v=\sqrt{2gh} [/tex]

Conversation of energy:

It states that the energy of interacting bodies or particles in a closed system remains constant.

It is given that:

A stone is dropped from rest at an initial height h above the surface of the earth.

To show:

The speed with which it strikes the ground

So, from conservation of Energy;

Total distance the stone will cover = [tex]h=\frac{1}{2} gt^2\\\\ [/tex]

The final velocity will be:

[tex]v=gt\\\\ v=g\sqrt{\frac{2h}{g} } \\\\ v=\sqrt{2gh} [/tex]

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