Question

when an object is dropped from the height of 100m from the surface of the earth,after 3 sec what will be distance travelled by the object.
(what will be the formula been used in this sum to find distance)

Answer 1

There is a simple equation for this that is worth remembering, if only because it comes up in high school physics exams, and if you can’t remember how to derive it, well, at least you can still use it.

[math]s = ut + \frac{1}{2}at^2[/math]

s = the distance travelled, which is what we are trying to find

u = is the initial velocity, which is 0.

t = the elapsed time, which is 4

a = acceleration, which is g, or 9.80665 m per second per second

since u is 0, we simply have

[math]s = \frac{1}{2}9.80665 \times 4^2 = 79.4532 metres[/math]

Note that the height itself isn’t a factor, as long as it’s greater than 79.45 metres of course.

This also takes no account of air resistance, which is usually the case with high school physics exam questions as well. To factor air resistance, we’d have to know a lot more about what object is being dropped, e.g. its mass and its drag coefficient.

Answer 2

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