Projectiles: Stone Thrown Vertically, Initial Velocity 19.6m/s

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The discussion revolves around a stone thrown vertically upwards with an initial velocity of 19.6 m/s. Participants are exploring the stone's velocity and position at various time intervals (1s, 2s, 3s) while grappling with the concepts of positive and negative values in the context of motion under gravity.

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  • Mixed

Approaches and Questions Raised

  • Participants share their calculations for position and velocity using kinematic equations, while some express confusion about the signs associated with upward and downward motion. Questions arise regarding the correctness of their answers and the interpretation of the equations.

Discussion Status

There is an ongoing exchange of calculations and clarifications, with some participants offering guidance on the correct application of formulas. Multiple interpretations of the problem are being explored, particularly concerning the direction of acceleration and the implications for velocity and position.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the amount of direct assistance they can receive. There is an emphasis on understanding the underlying physics concepts rather than simply obtaining answers.

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A stone is thrown vertically upwards with an initial velocity of 19.6m/s. What is it's velocity and position after 1s, 2s, 3s?
I'm getting confused between my positives and negatives. What answers did you get?
 
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A stone is thrown vertically upwards with an initial velocity of 19.6m/s. What is it's velocity and position after 1s, 2s, 3s?
I'm getting confused between my positives and negatives. What answers did you get?
First, write what you know:

Initial Velocity = 19.6m/s
acceleration = -9.81
Time = 1,2,3

So, from the formula d=Vi(t)+.5a(t)^2, we get:

di = 19.6(1) + (.5)(9.81)(1)^2
dii = 19.6(2) + (.5)(9.81)(2)^2
diii = 19.6(3) + (.5)(9.81)(3)^2

So:
di = 24.505m
dii = 58.82m
diii = 102.945m

And now, Vf^2 = Vi^2 + 2ad:

Vfi = Root[(19.6^2) + 2(-9.81)(24.505)]
Vfii = Root[(19.6^2) + 2(-9.81)(24.505)]
Vfiii = Root[(19.6^2) + 2(-9.81)(24.505)]

I don't have a calculator on hand, but these should work out to be the proper velocities, unless I messed up somewhere. :wink:
 
Last edited:
Quadratic said:
First, write what you know:
Initial Velocity = 19.6m/s
acceleration = -9.81
Time = 1,2,3
So, from the formula d=Vi(t)+.5a(t)^2, we get:
di = 19.6(1) + (.5)(9.81)(1)^2
dii = 19.6(2) + (.5)(9.81)(2)^2
diii = 19.6(3) + (.5)(9.81)(3)^2
So:
di = 24.505m
dii = 58.82m
diii = 102.945m
And now, Vf^2 = Vi^2 + 2ad:
Vfi = Root[(19.6^2) + 2(-9.81)(24.505)]
Vfii = Root[(19.6^2) + 2(-9.81)(24.505)]
Vfiii = Root[(19.6^2) + 2(-9.81)(24.505)]
I don't have a calculator on hand, but these should work out to be the proper velocities, unless I messed up somewhere. :wink:
Your answers are incorrect.

Initial velocity was upwards - gravity accelerates objects downward. Positive is upwards and negative is downwards. (In fact, you knew that. Unfortunately you didn't enter it into your equation that way.)

The velocity is just the derivative of the position equation:

[tex]v_f = v_i + at[/tex]

*acceleration is still downwards.
 
Yeah, I see what I did. Arithmetic always seems to kill me.
 
Bob G can you help me - getting so confused? What answers did you get because mine don't make sense
 
fabbo said:
Bob G can you help me - getting so confused? What answers did you get because mine don't make sense

Fabbo, sorry to bumb in like this but what have you done up till now. Please share us your results and how you got to them. Have you already consulted our marvellous tutorials ?

This is the best way to learn these important introductory concepts of Newtonian Physics.

regards
marlon
 
ok here goes...

s = ut + 1/2 at^2

so after 1 sec

s = 19.6 x 1 + 1/2 x 9.81 x 1^2
s = 14.7m

v = u + at

so after 1 sec

v = 19.6 + -9.81 x 2
v = 9.79 m/s

I did the rest like this and got answers for 2 seconds as - 0.02m/s and 19.58m and for 3 seconds - 9.83 m/s and 14.655m

I don't think they're right and I don't understand when the stone is going up and falling ....
 
fabbo said:
ok here goes...
s = ut + 1/2 at^2
Now we are getting started.
First of all, like you can see in the tutorial that i referred to, the above formula consists out of VECTORS. The clue is to express the components of these vectors with respect to the x and y-axis. For example, in this case, there is one acceleration : GRAVITY.
Acceleration is a vector and gravity is directed in the opposite direction (downward) of the y-axis. So you have
[tex]\vec{a_y} = -g \vec{e_y}[/tex]

The initial velocity is directed upwards in the same direction as the y-axis , so you have :
[tex]\vec{v_{initial}} = 19,6 \vec{e_y}[/tex]

WATCH THE SIGNS and be sure when to use + and - (same <--> opposite direction of the y-axis)

Now, try to write down the formula's for position and velocity in both the x and y direction.

The x-component of both acceleration and velocity are 0. Do you realize this ?

I strongly suggest you read the TUTORIAL. You will find more indept info on this along with some examples/exercises

regards
marlon

The general formula's for velocity (vector v) and position (vector r) are :

[tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]
[tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]
 
Last edited:
fabbo said:
ok here goes...
s = ut + 1/2 at^2
so after 1 sec
s = 19.6 x 1 + 1/2 x 9.81 x 1^2
s = 14.7m
v = u + at
so after 1 sec
v = 19.6 + -9.81 x 2
v = 9.79 m/s
I did the rest like this and got answers for 2 seconds as - 0.02m/s and 19.58m and for 3 seconds - 9.83 m/s and 14.655m
I don't think they're right and I don't understand when the stone is going up and falling ....
Yes, they are right (you're using the gravitational acceleration constant in your calculator, aren't you).

To find where the dividing line between going up and going down is, the second equation (v=u + at) should equal zero at that the precise instant between going up and coming down. If you know the initial velocity and the acceleration, you can find rearrange to find the time.
 
  • #10
BobG said:
Yes, they are right (you're using the gravitational acceleration constant in your calculator, aren't you).
To find where the dividing line between going up and going down is, the second equation (v=u + at) should equal zero at that the precise instant between going up and coming down. If you know the initial velocity and the acceleration, you can find rearrange to find the time.


No they are NOT correct. Look at the equation for the position, more specifically, look at the acceleration !


marlon
 
  • #11
does that make the first velocity 9.79m/s?

how does this fit into the position equation? i got 4.9m but was pretty confused...
 
  • #12
ok, how about this

v = 19.6 -9.81t
y = 19.6t -9.81t²/2

Try these...

marlon

i must say that you should be able to come up with these yourself.
 
  • #13
first velocity is indeed 9.79 m/s

first position 14.695 m
 
  • #14
oh ok i understand now, thank you for your help - it really is appreciated. I know to an experienced physicist I should be capable of doing these. But unfortunately I'm not and that is what is so great about this forum. I'm not looking for people to tell me answers, I just want to further my understanding. I have already asked my school teacher and I did not find he was able to help me. Therefore thank you and your tutorials are brilliant.
 
  • #15
fabbo said:
oh ok i understand now, thank you for your help - it really is appreciated. I know to an experienced physicist I should be capable of doing these. But unfortunately I'm not and that is what is so great about this forum. I'm not looking for people to tell me answers, I just want to further my understanding. I have already asked my school teacher and I did not find he was able to help me. Therefore thank you and your tutorials are brilliant.


if you have any other questions on this, please do not hesitate. I just want to stress (but i am sure many people have told you this) that it is very important that you know how to work with vectors/components, etc...

Good Luck

regards

marlon

edit : now i am off too, to see LOST on tv
 

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