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An object is released from rest at a height h. During the final second of its fall, it traverses a distance of 38m. Determine the value of h.

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AL2006

During the final second of its fall, it falls a distance of 38m.

Its average speed during that final second was 38 m/s.

But we know that it gained 9.8 m/s of speed during that second.

Its speed at the end of that second must have been (38+4.9) = 42.9 m/s ,
and at the beginning of that second must have been (38-4.9) = 33.1 m/s .

Since its speed at the beginning of the final second was  33.1 m/s,
that final second began at  (33.1/9.8) = 3.378 seconds after the drop.

All together, when the final second is added onto that, the object
fell for a total of  4.378 seconds.

  Distance of fall from rest = (1/2) (g) (t)²

                                      = (4.9 m/s²) (4.378 s)²

                                      = (4.9 m/s²) (19.163 s²)

                                      =      93.9  meters  .

nuhulawal20

The height \(h\) from which the object was released from is 93.9m.

Since object is released from rest

Initial velocity; \(u = 0m/s\)

First, we find the velocity \(v\) of the object at the beginning of the last second.

During the final second of its fall, it traverses a distance of 38m

Distance travelled \(h_1 = 38m\) ( final second; time \(t = 1s\) )

Using the second equation of motion:

\(s = ut + \frac{1}{2}at^2\)

We substitute in our values

\(38m = (v * 1s) + ( \frac{1}{2} * 9.8m/s^2 * (1s)^2 )\\\\38m = (v * 1s) + 4.9m\\\\(v * 1s) = 38m - 4.9 m\\\\(v * 1s) = 33.1m\\\\v = 33.1m/s\)

Now Lets find the height \(h\) as the object transverse, final velocity \(v = 33.1m/s\), initial velocity \(u = 0m/s\) and acceleration due to gravity; \(a = 9.8m/s^2\)

From the Third Equation of Motion:

\(v^2 = u^2 + 2ah\)

We substitute in our values

\((33.1m/s)^2 = 0m/s + (2 * 9.8m/s^2 * h )\\\\1095.61m^2/s^2 = 19.6m/s^2 * h \\\\h = \frac{1095.61m^2/s^2}{19.6m/s^2}\\\\h = 55.9m\)

So \(h_2 = 55.9m\)

To get the height from which the ball was released:

Height = \(h_1 + h_2\)

Height = \(38m + 55.9m\)

Height = \(93.9m\)

Therefore, the height \(h\) from which the object was released from is 93.9m.

Learn more; https://brainly.com/question/14232141

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