The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall occur on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?

Answer :

elcharly64

Answer:

The object could fall from six times the original height and still be safe

Explanation:

Free Falling

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

[tex]U=m.g.h[/tex]

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

[tex]\displaystyle K=\frac{1}{2}m.v^2[/tex]

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

[tex]\displaystyle m.g.h=\frac{1}{2}m.v^2[/tex]

Solving for h

[tex]\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}[/tex]

If the drop had occurred in the Moon, then

[tex]\displaystyle h_M=\frac{v_M^2}{2g_M}[/tex]

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

[tex]\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}[/tex]

[tex]\displaystyle h_M=6\frac{v^2}{2g}=6h[/tex]

This means the object could fall from six times the original height and still be safe

The object could fall from six times the original height and still be safe.

What is free falling?

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E) of

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

k.E = 1/2mv²

where v is the speed just before hitting the ground.

From energy conservation principle, initial and final mechanical energy are equal.

mgh =  1/2mv²

The height from which one could fall is

h = v²/2g

If the drop had occurred in the Moon, then

h(M) = v(M)²/2g(M)

The gravitational acceleration on the Moon is g(M)=1/6 g(E) gravitational acceleration on the Earth.

h(M) = v(M)²/2(g/6)

h(M) =6 [ v(M)²/2g]

Thus, the object could fall from six times the original height and still be safe

Learn more about free falling.

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