Vertical Component of a ball rolling off table

AI Thread Summary
The discussion addresses a physics problem involving a ball rolling off a table with a horizontal velocity of 5 m/s. The vertical component of the ball's velocity just before it hits the floor is calculated using the formula V = Vo + at, resulting in a downward velocity of 6 m/s. The horizontal component remains constant at 5 m/s since no horizontal forces act on the ball during its fall. Participants clarify that the initial vertical velocity is zero and that gravity solely influences the vertical motion. The final answers are confirmed as 6 m/s downward for the vertical component and 5 m/s for the horizontal component.
Chuck Norris
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Homework Statement


A ball rolls off a table with a horizontal velocity of 5 m/s. If it takes 0.6 seconds for it to reach the floor what is the vertical component of the ball's velocity just before it hits the floor? (use g= 10 m/s^2)

The second part of the question is what is the horizontal componenet ofthe ball's velocity just before it hits the floor?

Homework Equations



D vert: 1/2at^2

The Attempt at a Solution



I'm not sure if this is correct:

Dvert = 1/2at^2
1/2(10m/s^2)(0.6s)^2
Dvert= 1.8m

For the second part, would I do the exact same thing but plug in different times to plot the different spots the ball would project at on the way down?

Thanks for your help.
 
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Chuck Norris said:
I'm not sure if this is correct:

Dvert = 1/2at^2
1/2(10m/s^2)(0.6s)^2
Dvert= 1.8m
You're supposed to find the velocity, not the distance fallen.
 
1] Your answer to A is in m, yet you are asked for m/s.

2] Seems to me the answers are pretty simple. The vertical component and horizontal component are completely independent of each other. It would seem to me that the horizontal component is trivial (almost a trick question) - it does not even need to be calculated.
 
So then am I using the wrong formula all together?
 
Chuck Norris said:
So then am I using the wrong formula all together?
Yep. You're using a formula for distance, but you need the formula for velocity.
 
V = V0 + at

V= 5 m/s + (10m/s^2)(0.6s)

V= 5.6 m/s?

I'm sure this is the wrong formula also.
 
Chuck Norris said:
V = V0 + at

V= 5 m/s + (10m/s^2)(0.6s)

V= 5.6 m/s?

I'm sure this is the wrong formula also.
This is the right formula, but you are using the wrong initial velocity. (5 m/s is the horizontal component of the initial velocity, not the vertical.) Treat vertical motion separately from horizontal. (Also: Double check your arithmetic.)
 
V = Vo + at

V= 0 + (10 m/s^2)(0.6s)

V= 6 m/s

I'm assuming since there is not a initial velocity given in the problem that the initial velocity is assumed to be 0. Sorry for all the dumb questions, physics has just been rough for me so far. Thanks for all your help so far.
 
Chuck Norris said:
V = Vo + at

V= 0 + (10 m/s^2)(0.6s)

V= 6 m/s
Perfect. The vertical component is 6 m/s downward. (So they might want you to call it - 6 m/s.)
 
  • #10
Awesome! Thank you so much. Now for the second part,
what is the horizontal componenet ofthe ball's velocity just before it hits the floor?

Would I use the formula d(horiziontal)=Vot?

Now I'm confused because if I use that formula and plug in the initial velocity of 0 obviously I'm going to get an answer of 0.
 
  • #11
Chuck Norris said:
Awesome! Thank you so much. Now for the second part,
what is the horizontal componenet ofthe ball's velocity just before it hits the floor?

Would I use the formula d(horiziontal)=Vot?

Now I'm confused because if I use that formula and plug in the initial velocity of 0 obviously I'm going to get an answer of 0.

What is the initial horizontal velocity of the ball? What forces are acting on the ball horizontally (i.e. acceleration)?
 
  • #12
DaveC426913 said:
What is the initial horizontal velocity of the ball? What forces are acting on the ball horizontally (i.e. acceleration)?

The intial horizontal velocity is 5 m/s when it rolls off the table.

So would it be:

d=Vot

D= (5m/s)(0.6s)

d= 3m?
 
  • #13
nevermind i see that I'm looking for velocity again yet i went back to answering in distance
 
  • #14
Would I just use the same formula as I used with the vertical component but plug in the 5m/s as the initial velocity instead of 0?
 
  • #15
Haha wait so is the answer just 5 m/s? Because horizontal velocity doesn't change?
 
  • #16
You're overthinking it. What is the initial horizontal v? What force(s) are acting to change the horizontal v?
 
  • #17
The initial horizontal velocity is 5 m/s correct? I don't believe any forces are acting to change it...
 
  • #18
unless the initial horizontal velocity is just zero...since there are no forces changing it
 
  • #19
Chuck Norris said:
The initial horizontal velocity is 5 m/s correct? I don't believe any forces are acting to change it...
Correct and correct.

So...
 
  • #20
The answer is just 5 m/s.
 
  • #21
maybe?
 
  • #22
Chuck Norris said:
The answer is just 5 m/s.
Right. The horizontal velocity component starts out at 5 m/s and stays at 5 m/s. Gravity only affects the vertical component.
 
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