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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that is a distance d from the point of release. If this distance was quadrupled, what would the new speed be?

Respuesta :

Answer:

When an object is released from rest and falls freely under the influence of gravity, its speed at any point can be calculated using the equation for the final speed of an object in free fall:

v = √(2gh)

Where:

v = final speed

g = acceleration due to gravity (approximately 9.81 m/s² near the surface of the Earth)

h = height from which the object is released

If the distance d is quadrupled to 4d, the new speed can be calculated using the same equation:

v' = √(2g(4d))

Now, we can simplify the equation:

v' = √(8gd)

Since 8gd is 4 times greater than 2gd, the new speed v' will be the square root of 4 times the original speed v.

Therefore, the new speed will be twice the original speed:

v' = 2v

Explanation:

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