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Trava [24]
3 years ago
8

A ball droped from a building. How fast is it traveling after falling 3.55s

Physics
1 answer:
algol [13]3 years ago
4 0

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

d=ut+\dfrac{1}{2}at^2

Here, u = 0 and a =g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m

So, it will cover 61.75 m after falling 3.55 seconds.

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Answer:

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Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

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Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

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