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Set 4 Problem number 13
An object falls freely from rest in the
vicinity of the earth's surface.
- 4.9 seconds later, how fast will it be moving and
how far will it have fallen?
In the vicinity of the Earth's surface, an object will accelerate at 9.8
m/s.
- In 4.9 seconds, an object accelerating from rest at 9.8 m/s will attain a
speed of 48.02 m/s.
- Its average speed will be (0+ 48.02)/2 m/s = 24.01 m/s.
- Thus in 4.9 seconds it will ahve traveled ( 24.01 m/s)( 4.9 sec)= 117.649 m/s.
In a constant acceleration situation, change in velocity is equal to the product
of acceleration and time interval.
- It follows that in a time interval of duration `dt an object accelerating under
the influence of constant gravitational acceleration g will change its velocity by
- Starting from rest the velocity attained will be equal to the change `dv = g `dt.
- The object will have an average velocity of
- vAve = (0 + g `dt) / 2 = .5 g `dt during its fall.
- It will therefore have traveled distance
- `ds = vAve * `dt = (.5 g `dt) * `dt = .5 g `dt^2 during its fall.
This last result could have been obtained directly from the equation `ds = v0
`dt + .5 a `dt^2, which applies to uniformly accelerated motion:
- For the present situation v0 = 0 and a = g, so `ds = .5 g `dt^2.
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