Vertical Component of a ball rolling off table

[Chuck Norris]

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.

 

[Doc Al]

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.

 

[DaveC426913]

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.

 

[Chuck Norris]

So then am I using the wrong formula all together?

 

[Doc Al]

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.

 

[Chuck Norris]

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.

 

[Doc Al]

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.)

 

[Chuck Norris]

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.

 

[Doc Al]

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.)

 

[Chuck Norris]

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.

 

[DaveC426913]

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)?

 

[Chuck Norris]

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?

 

[Chuck Norris]

nevermind i see that I'm looking for velocity again yet i went back to answering in distance

 

[Chuck Norris]

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?

 

[Chuck Norris]

Haha wait so is the answer just 5 m/s? Because horizontal velocity doesn't change?

 

[DaveC426913]

You're overthinking it. What is the initial horizontal v? What force(s) are acting to change the horizontal v?

 

[Chuck Norris]

The initial horizontal velocity is 5 m/s correct? I don't believe any forces are acting to change it...

 

[Chuck Norris]

unless the initial horizontal velocity is just zero...since there are no forces changing it

 

[DaveC426913]

The initial horizontal velocity is 5 m/s correct? I don't believe any forces are acting to change it...

Correct and correct.

So...

 

[Chuck Norris]

The answer is just 5 m/s.

 

[Chuck Norris]

maybe?

 

[Doc Al]

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.