A 1kg mass is dropped from a height of 40m. At what height will the velocity be 1/4 of the velocity attained before impact?

To find at what height the velocity will be 1/4 of the velocity attained before impact, we need to calculate the velocity at impact first.

We can use the equation of motion for an object in free fall:

v² = u² + 2as

Where:
v = Final velocity (at impact)
u = Initial velocity (0 m/s in this case since it is dropped)
a = Acceleration due to gravity (-9.8 m/s², taking downward as negative)
s = Displacement or height (40 m)

Plugging in the values into the equation:

v² = 0² + 2 * (-9.8) * 40
v² = -784
v = √(-784)
v = -28 m/s

Since the velocity is negative, indicating it is moving downward, we can consider its magnitude for comparison purposes.

Now, let's find the height at which the velocity is 1/4 of the velocity at impact.

We can use the same equation of motion, but this time v will be 1/4 of the velocity at impact, which is (-28 m/s in magnitude):

v = √(2as)

Plugging the values into the equation:

-28 = √(2 * (-9.8) * s)
(-28)² = 2 * (-9.8) * s
784 = -19.6s
s = 784 / (-19.6)
s ≈ -40 m

The negative sign indicates that the height is below the initial position. However, since heights are usually represented as positive values, we can consider the magnitude and state that the height at which the velocity will be 1/4 of the velocity at impact is approximately 40 meters below the starting point.