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Answer:
Horizontal distance between the log and the bridge when the stone is released = 17.24 m
Step-by-step explanation:
Height of bridge, h = 50.8 m
Speed of log = 5.36 m/s
We need to find the horizontal distance between the log and the bridge when the stone is released, for that first we need to find time taken by the stone to reach on top of log,
We have equation of motion. s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50.8 m
Substituting,
s = ut + 0.5 at²
50.8 = 0.5 x 9.81 x t²
t = 3.22 seconds,
So log travels 3.22 seconds at a speed of 5.36 m/s after the release of stone,
We have equation of motion. s = ut + 0.5 at²
Initial velocity, u = 5.36 m/s
Acceleration, a = 0 m/s²
Time, t = 3.22 s
Substituting,
s = ut + 0.5 at²
s = 5.36 x 3.22 + 0.5 x 0 x 3.22²
s = 17.24 m
Horizontal distance between the log and the bridge when the stone is released = 17.24 m
The horizontal distance between the log and the bridge when the stone is released is 17.26 meters approx.
What are equations of motion?
There are three equation of motions that can be used when motion of the object is under constant acceleration and on a linear path.
They are listed below as:
[tex]v = u + at\\\\s = ut + \dfrac{1}{2} at^2\\\\v^2 = u^2 + 2as[/tex]
where the symbols have following meanings:
- u = initial velocity of the considered object
- v =- final velocity of the object
- a = acceleration of the object in the direction of motion
- s = displacement of the object in 't' time.
The stone is dropped down from the height of 50.8 meters. It is dropped down with 0 as initial velocity. The acceleration applied is the gravitational acceleration(for earth).
Let the stone takes 't' seconds to travel to the surface of the log(assuming it is 50.8 meters down).
Therefore, we have:
- Time taken for travel = t
- S = displacement of object traveled = -50.8 meters
- u = 0 m/s
- a = -g = -9.8 m/sq. seconds.
From second equation of motion, we get:
[tex]-50.8 = 0\times t + \dfrac{1}{2} \times -9.8 \times t^2\\t = \sqrt{\dfrac{50.8}{4.9}} \approx 3.22 \: \rm sec[/tex]
Since the speed of the log was 5.36 m/s, thus, in 3.22 seconds the log was moved about [tex]5.36 \times 3.22 \approx 17.26 \: \rm meters[/tex]
This is the distance (horizontal) between the log's initial position and the bridge.
Thus, the horizontal distance between the log and the bridge when the stone is released is 17.26 meters approx.
Learn more about equation of motions here:
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