B Drop Ball Moving: Initial Velocity 0m/s?

AI Thread Summary
When dropping a ball while walking at 2 m/s, the initial vertical velocity of the ball is 0 m/s at the moment of release, but it retains the horizontal velocity of 2 m/s. The discussion clarifies that the concept of "dropping" refers specifically to vertical motion, while horizontal motion is independent. Therefore, the ball will follow a parabolic trajectory due to the combination of these two motions. Understanding the independence of horizontal and vertical components is key to solving related physics problems. The conversation emphasizes the importance of vector analysis in comprehending motion dynamics.
elf197320501
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Hi I'm trying to solve a problem but I'm a bit confused about this part:

Suppose you are walking at a speed of 2 m/s when you drop the ball that you were holding in your hands.

Is the initial velocity 0 m/s? In the textbook it says that initial velocity for dropped objects is 0 m/s but that was for dropping it while stationary, does the same law apply when an object is dropped while moving? thanks!
 
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Hi ya
welcome to PF :smile:

elf197320501 said:
Is the initial velocity 0 m/s? In the textbook it says that initial velocity for dropped objects is 0 m/s but that was for dropping it while stationary,

that's correct

elf197320501 said:
does the same law apply when an object is dropped while moving? thanks!

think about what is happening to the ball ... you have already stated that it has an initial velocity ... just not downwards
so what do you think is going to happen to the ball when you drop it when there is horizontal motion ?
what do you think its path towards the ground going to look like ?Dave
 
elf197320501 said:
Hi I'm trying to solve a problem but I'm a bit confused about this part:

Suppose you are walking at a speed of 2 m/s when you drop the ball that you were holding in your hands.

Is the initial velocity 0 m/s? In the textbook it says that initial velocity for dropped objects is 0 m/s but that was for dropping it while stationary, does the same law apply when an object is dropped while moving? thanks!

What if you don't drop the ball? At the last second you decide not to let it go. What's its initial velocity then?

What if you just about let it go - it's almost out of your hand - maybe you have just about let it go, but your hand is still very near and then you get hold of it again?

If it's not moving forward when you drop it, then at exactly what point in the "letting go" process does it stop moving forward?
 
Perhaps read up on vectors and how a velocity vector can be made up from, or treated as, the sum of two others eg horizontal and vertical component vectors.
 
The inital velocity in the downwards direction(ill state it as the y axis) would be 0, wouldn't it? Where as the inital velocity of the ball in the direction which you are moving(x axis) after the ball left your hand would be 2m/s, i think. Are you supposed to determine when the ball hits the ground? If so, convert the vectors of the ball on each axis to their polar coordinates. I think?
 
No need for polar coordinates. Try doing the experiment in a train or on a rotating planet. Does the horizontal motion affect the time to fall vertically at all? How long does a bullet fired horizontally from 1m take to hit the ground?
 
The ball has both horizontal and vertical motion. These can be treated as independent -- that is, you can ignore the horizontal motion when you are calculating the vertical motion and vice versa. What is the horizontal motion of the ball? What direction are the forces on the ball?
 
elf197320501 said:
Hi I'm trying to solve a problem but I'm a bit confused about this part:

Suppose you are walking at a speed of 2 m/s when you drop the ball that you were holding in your hands.

Is the initial velocity 0 m/s?

The initial velocity of what ? and in what context.

In the textbook it says that initial velocity for dropped objects is 0 m/s

That's partly because velocity is a speed that includes direction ("55mph, traveling due West"), whether stated or implied. Semi-stated in the textbook is that the 0m/s refers to the "drop"

And it's partly because textbook authors muddle things up a bit. Obviously if you're walking and let go of something, it's going to keep the speed and velocity of the walk at first. In this case you can be pretty sure that "drop" as a verb is being used to mean "drop in altitude", where "altitude" refers to a vertical displacement from the ground.

but that was for dropping it while stationary, does the same law apply when an object is dropped while moving? thanks!

As above. The initial velocity is a "0m/s drop". Depending on what the actual homework question is they may or may not want you to include the "2m/s horizontal" walk as well.
 
hmmm27 said:
That's partly because velocity is a speed that includes direction ("55mph, traveling due West"), whether stated or implied. Semi-stated in the textbook is that the 0m/s refers to the "drop"

no, it was stated as being stationary, no muddles anywhereAnyway guys

The OP @elf197320501 hasn't bothered to return to let us all know if his Q was answered in a way he could understand or not
we may never know if he tried doing the vectors for the problem

Dave
 
  • #10
davenn said:
no, it was stated as being stationary, no muddles anywhere

no, the "stationary" qualifier was added by the poster.

The muddle is the assumption of the definition of "drop", made by herself or the textbook.
 
  • #11
hmmm27 said:
no, the "stationary" qualifier was added by the poster.

read it again ... he/she stated the TEXTBOOK reference

In the textbook it says that initial velocity for dropped objects is 0 m/s but that was for dropping it while stationary,...
 
  • #12
well yes, that's what I read, too. The "but that was for..." bit was obviously added by the poster; it could be an interpretation of what the textbook explicitly says, or it could be a deduction based on common sense.

My "she" assumption is sexist, based on the "elf" username, in ignorance of whether the site automatically generates a user name if none is provided, or not.
 
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  • #13
Thank you for all the replies, I think I understand and can solve the problem now that I know horizontal and vertical motion are independent (that was mainly what was causing my confusion)

and this is the exact wording from the textbook by the way: "Whenever we say "dropped", it means Vo= 0."
 
  • #14
Glad we could help.
CWatters said:
Perhaps read up on vectors and how a velocity vector can be made up from, or treated as, the sum of two others eg horizontal and vertical component vectors.
 
  • #15
CWatters said:
Perhaps read up on vectors and how a velocity vector can be made up from, or treated as, the sum of two others eg horizontal and vertical component vectors.
Many quantities can be represented by Vectors (with magnitude and direction) and velocity (motion) happens to be one of them. To get an idea of how vectors work, it could be worth while thinking in terms of forces (another example of vectors). It's perhaps a bit more concrete and tangible to think about forces, to start with. We started at school with 'The Parallelogram of Forces' (look it up), which tells you what the resultant of two forces, acting in two different directions. It is fairly obvious that a combination of a force, sideways and a force upwards will result in an effective diagonal force. You can even do simple home experiments on a table top, dragging things with two strings in different directions. All the same rules apply to adding velocities (with things on the move) but it is not always as intuitive as what happens with forces which can be static.
 
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