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Set 4 Problem number 13
How fast will a freely falling object the moving 2.4 seconds after beginning its free
fall from rest?
How far will it have fallen during this time?
In the vicinity of the Earth's surface, an object will accelerate at 9.8
m/s.
- In 2.4 seconds, an object accelerating from rest at 9.8 m/s will attain a
speed of 23.52 m/s.
- Its average speed will be (0+ 23.52)/2 m/s = 11.76 m/s.
- Thus in 2.4 seconds it will ahve traveled ( 11.76 m/s)( 2.4 sec)= 28.224 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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