## Help with a problem?

##### what school do you go to?

lolcatslol
Posts: 402
Joined: Wed May 22, 2013 4:44 pm

### Help with a problem?

I'm trying to learn physics this year, and I got stuck on a problem.

A baseball is seen to pass upward by a window 23 m above the street with a vertical speed of 9 m/s. The ball was thrown from the street.
(a) What was its initial speed?
23.06 m/s

(b) What altitude does it reach?
27.13

(c) How long after it was thrown did it pass the window?
1.43

(d) After how many more seconds does it reach the street again?

Any help is appreciated Last edited by lolcatslol on Tue Jun 11, 2013 8:23 pm, edited 1 time in total.

tomwzhere
Posts: 1922
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### Re: Help with a problem?

I just finished physics, which although was one of the worst classes I have ever taken, I'm surprised I learned a lot with such a horrible teacher. For your problem locate the horizontal and vertical acceleration and velocity.
So, since you know this is a parabolic motion, you know that at the peak of its altitude it's velocity becomes 0 before going back down. You then substitute the 0 m/s into the v(final) and the initial from part a into the equation v(final)=v(initial)+at. Solve for time and after substitute the known values into the equation x=vt+1/2at^2. (Note that the acceleration is negative since the force of gravity is pointing downwards).
For c, you know that the speed when it passes the window is 9 m/s. So use the equation v(final)=v(initial)+at. 9 m/s would be your v(final) and you know your initial from part a.
The time you get from part b to reach the highest altitude should be half of the time it takes to get from street to street, since there are no other external forces acting on it besides the constant gravity. So, just multiply the time acquired in part b by 2 and that is the time it takes to reach the street again.

CaptainJamesTKirk
Posts: 2374
Joined: Thu Nov 12, 2009 8:10 pm
Location: Anahiem
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### Re: Help with a problem?

Sorry cat I'm not much help this is Spocks department. . But with the help of google I found this:

What altitude does it reach?
Y=Yo + Vot+ 1/2at^2

Y is maximum height
Yo is initial vertical placement
Vo is velocity
a is always equal to g which is in this case -9.8m/s^2
t is time needed to reach point
Hope it helps.
and this http://www.physicstutorials.org/home/me ... /free-fall" onclick="window.open(this.href);return false; lolcatslol
Posts: 402
Joined: Wed May 22, 2013 4:44 pm

### Re: Help with a problem?

tomwzhere wrote:I just finished physics, which although was one of the worst classes I have ever taken, I'm surprised I learned a lot with such a horrible teacher. For your problem locate the horizontal and vertical acceleration and velocity.
So, since you know this is a parabolic motion, you know that at the peak of its altitude it's velocity becomes 0 before going back down. You then substitute the 0 m/s into the v(final) and the initial from part a into the equation v(final)=v(initial)+at. Solve for time and after substitute the known values into the equation x=vt+1/2at^2. (Note that the acceleration is negative since the force of gravity is pointing downwards).
For c, you know that the speed when it passes the window is 9 m/s. So use the equation v(final)=v(initial)+at. 9 m/s would be your v(final) and you know your initial from part a.
The time you get from part b to reach the highest altitude should be half of the time it takes to get from street to street, since there are no other external forces acting on it besides the constant gravity. So, just multiply the time acquired in part b by 2 and that is the time it takes to reach the street again.
thanks i got that.
the only problem is that I'm now having problem with (d)

I timed what i got in (b) by 2, but i still got the wrong answer.

/e
for part (d)
I am going to do the speed from (a) divided by 9.8 (2.35) multiply it by 2 ( 4.70) and then subtract what I have for (c) (3.27)
does that seem logical?

/ee it worked.
Thanks you guys. I might need more help later though ### Who is online

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