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The velocity of a free-falling object is given by [tex]v = \sqrt{2gh}[/tex], where:

- [tex]v[/tex] is the velocity in meters per second (m/s),
- [tex]g[/tex] is the acceleration due to gravity in meters per second squared (m/s²),
- [tex]h[/tex] is the distance fallen in meters (m).

If the object hits the surface with a velocity of 30 m/s and the acceleration due to gravity on the moon is [tex]g = 1.57 \, \text{m/s}^2[/tex], determine the height from which the object was dropped.

Answer :

The object was initially dropped from a height of about 287.6 meters.

We can change the acceleration due to gravity on the moon to g = 1.57 [tex]m/s^2[/tex] and use this value in the formula:

h = [tex]900 m^2/s^2[/tex] / (2 × 1.57 m/s^2)

h ≈ 287.6 m

what is velocity?

The pace at which an object's position changes in relation to a frame of reference and time is what is meant by velocity. Although it may appear sophisticated, velocity is just the act of moving quickly in one direction. Since it is a vector quantity, the definition of velocity requires both magnitude (speed) and direction. Its SI equivalent is the meter per second (ms-1). A body is considered to be accelerating if its velocity changes, either in magnitude or direction.

from the question:

The starting height from which the object was dropped can be determined using the formula for the velocity of a free-falling object. When an object strikes a surface at 30 m/s, we can write:

30 m/s = √(2gh)

Squaring both sides gives:

[tex]900 m^2/s^2[/tex] = 2gh

Solving for h:

h = [tex]900 m^2/s^2[/tex]/ (2g)

We can change the acceleration due to gravity on the moon to g = 1.57 [tex]m/s^2[/tex] and use this value in the formula:

h = [tex]900 m^2/s^2[/tex] / (2 × 1.57 m/s^2)

h ≈ 287.6 m

As a result, the object was dropped from a height of roughly 287.6 meters.

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