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Set 4 Problem number 13
An object begins free fall from rest.
- How fast will it be moving after 5.3 seconds of
free fall?
- How far will it fall during the 5.3 seconds?
In the vicinity of the Earth's surface, an object will accelerate at 9.8
m/s.
- In 5.3 seconds, an object accelerating from rest at 9.8 m/s will attain a
speed of 51.94 m/s.
- Its average speed will be (0+ 51.94)/2 m/s = 25.97 m/s.
- Thus in 5.3 seconds it will ahve traveled ( 25.97 m/s)( 5.3 sec)= 137.641 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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