Physics, 19.10.2019 13:10 heavenwagner

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.

Answers: 1

Show answers

Answers

Answer from: dnation94

The height 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 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 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 from which the object was released from is 93.9m.

Learn more; link

Answer from: Michael9979

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 from: Quest

an electric circuit

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia