Question

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.

Answer 1

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  .

Answer 2

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.

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