Physics People HELP me out.

[jcovercash]

I want to make sure I am doing this correctly That way my whole lab is not screwed up.

A ball is throw a distance of 60 ft. (18.24 m). It takes the Ball 1.11 s to travel this distance.

I need to find the following.

Delta Y
VIx
VIy
VI
Angle the ball was thrown.

I came up with
Delta Y--.00271 m OR 3.02?????
VIx--16.43 m
VIy--5.439
VI--17.307 (m/s)
Angle the ball was thrown--18.317 degrees

LMK if this sounds right to you.

Thanks,
Josh

 

[jcovercash]

ANYONE?????????//

 

[Nitemare]

Is it just me or does Physics bite the big one?


Wish I could help, but I just started Intro to College Physics a week or 2 ago

 

[gopunk]

Originally posted by: Nitemare
Is it just me or does Physics bite the big one?


Wish I could help, but I just started Intro to College Physics a week or 2 ago

it's better (easier) than chemistry... *shudder*

 

[Haircut]

Looks OK to me.

Assuming you are talking about the ball moving 18.24m along the x-axis and not the ball actually travelling 18.24m along its curved trajectory.

t=1.11.

Ymax occurs at t=0.555s and Vy = 0 at this point

Vy = Viy - gt.
Assume g=9.8 m/s/s

0= Viy - 5.439, thus Viy = 5.439 m/s

x = Vx * t, assuming no wind resistance.
18.24 = Vx * 1.11
Vx = 16.43 m/s

Using Pythagoras you get Vi = 17.31 m/s and you can use simple trig on Vx and Viy to get the angle to be 18.31 degrees.

I don't know what you mean when you say to calculate delta y though, y is a function of t so an actual numerical value will change over time.

 

[gopunk]

I don't know what you mean when you say to calculate delta y though, y is a function of t so an actual numerical value will change over time.

maybe i'm missing something, but isn't that the point of delta Y? to calculate the change? (which i guess is 0... max delta Y is what you want?)

 

[Legendary]

delta y, in this case i believe, is the maximum change in y (maximum height)

I believe the above calculations to be correct.

 

[jcovercash]

Originally posted by: Haircut
Looks OK to me.

Assuming you are talking about the ball moving 18.24m along the x-axis and not the ball actually travelling 18.24m along its curved trajectory.

t=1.11.

Ymax occurs at t=0.555s and Vy = 0 at this point

Vy = Viy - gt.
Assume g=9.8 m/s/s

0= Viy - 5.439, thus Viy = 5.439 m/s

x = Vx * t, assuming no wind resistance.
18.24 = Vx * 1.11
Vx = 16.43 m/s

Using Pythagoras you get Vi = 17.31 m/s and you can use simple trig on Vx and Viy to get the angle to be 18.31 degrees.

I don't know what you mean when you say to calculate delta y though, y is a function of t so an actual numerical value will change over time.

Ok thanks that helped me. Thats all I needed. I just started 2 weeks ago, and didnt sign up for the class. they just gave it to me. I wanted something easy for my last year in high school for science but I got physics. I figured that since I was in it I might as well take it. But physics is easier than chemistry,

Josh

 

[Haircut]

Originally posted by: gopunk
I don't know what you mean when you say to calculate delta y though, y is a function of t so an actual numerical value will change over time.

maybe i'm missing something, but isn't that the point of delta Y? to calculate the change? (which i guess is 0... max delta Y is what you want?)

Well, it was confusing me a little when he posted that delta Y was 0.00271m or 3.02m.
It would be easy to calculate y as a function of time, I just wasn't sure what he wanted with posting that above.

 

[jcovercash]

Originally posted by: Haircut

Originally posted by: gopunk
I don't know what you mean when you say to calculate delta y though, y is a function of t so an actual numerical value will change over time.

maybe i'm missing something, but isn't that the point of delta Y? to calculate the change? (which i guess is 0... max delta Y is what you want?)

Well, it was confusing me a little when he posted that delta Y was 0.00271m or 3.02m.
It would be easy to calculate y as a function of time, I just wasn't sure what he wanted with posting that above.

Yea sorry about that. I messed up and caught my mistake and added that to make sure it was a mistake. delta Y is the max height that is goes not the changed in total height. So you take the time /2 and use the first half of the equation or so my teacher says.

Josh

 

[Keego]

I just finished physics w/ calculus (college) and I will *NEVER* do another physics problem again. Not even this one

 

[Haircut]

If you are talking about the maxium height the ball reaches then I don't get either of the values you posted.

You will have to use the equation s= ut+ (1/2)at^2, where s= distance, u = initial speed and a = acceleration to calculate the distance travelled in the y direction at time t.
Substitute the values in and you should get the correct answer.