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An object of mass 2.00 kg is held at a position A, a vertical height of 20.0 m above the ground. Point B is 8.00 m directly below A. Neglect air resistance and use g = 10.0 m/s2. What is the speed of the object at position B when it falls from position A? I have no idea how to go about this...

2 Answers

4 votes
find the time for the object to fall from position A to position B. You can use \[x = \frac{1}{2}at^2\]to do so. \(x\) represents the distance traveled, \(a\) is the acceleration (here due to gravity, so use the value provided for \(g\)), and \(t\) is the time. When you have found the time required for the object to fall from position A to position B, you can use \[v = at\] to calculate the speed. Again, \(a=g\) and \(t\) is the value you found in the first part!!!! ;)
User Othmane
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6 votes

There is a lot of smoke and cob-webs in this problem, put there
to distract people who aren't quite sure of what they're doing.

First of all, the mass of the object is irrelevant. If there's no air resistance,
then everything falls at the same rate ... feathers, rocks, cars, battleships ...
everything !

Next, the 20m and the 8m above the ground aren't necessary. All that really
matters is that the object is dropped, and we want to know its speed after it
falls 12 meters.

The distance covered by an accelerating object in 't' seconds is

Distance = (1/2) (acceleration) (t²)

In gravity, 12 meters = (1/2) (10) (t²)

Divide each side by 5: 12/5 = t²

Square root each side: t = √2.4 = 1.55 seconds (rounded)

That's the time it takes an object to fall 12 meters after you drop it . . .
1.55 seconds. Now the question wants to know: How fast is it falling
at that time ?

The 'acceleration of gravity' means how much the speed of a falling object
grows each second. The question told us to use 10m/s² for gravity, so that's
what we'll use. It's pretty close to the real number. It means that when an
object is falling, its speed grows by 10m/s every second ... every second,
it's falling 10 m/s faster than it was falling 1 second earlier.

After it falls for 1.55 second, its speed is (1.55s) x (10 m/s²) = 15.5 m/s faster
than at the beginning of the 1.55 second.

What was its speed at the beginning of the 1.55 sec ? It was originally held
in somebody's hand and then dropped, so its original speed was zero. After
1.55 sec, it has fallen 12 meters, and its speed is 15.5 m/s.

User BertR
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7.5k points