Question:
A baseball is hit almost straight up into the air with a speed of about 41 m/s.
(a) How high does it go? (m)
(b) How long is it in the air? (s)
I think you probably just have to multiply or divide this number (41 m/s) by some sort of wind resistance ratio or number, but I have no idea what this would be since I've never had physics before and my teacher insists we don't need a book for this class.
Please help give me the tools and I can figure it out.
Defennder
Aug25-08, 11:19 PM
a)You have the standard 3 kinematics equations, right? Use them. Note that the only acceleration here is the one due to gravity, which is constant.
b)Same approach as before.
EDIT:You want to consider "wind"? That becomes difficult, especially if the wind is blowing upwards or downwards. You didn't provide any details as well.
wadesweatt
Aug25-08, 11:21 PM
Like I said, I have no book and I am a week into (one class so far) my first physics class ever. So, NO I don't have kinematics equations or any idea what that is.
Defennder
Aug25-08, 11:24 PM
How would you do this without any equations? You may not need a textbook but you would surely need notes. If you're not given the equations, then you can look them up online on Google.
wadesweatt
Aug25-08, 11:29 PM
How would you do this without any equations?.
I guess that's why I'm here. Because I couldn't do it.
I looked up the equations already. They call for acceleration, time, displacement, and velocity inputs on different variables.
I guess I am looking for displacement, but I am only given velocity. What do I do about the other two variables?
LowlyPion
Aug25-08, 11:32 PM
Like I said, I have no book and I am a week into (one class so far) my first physics class ever. So, NO I don't have kinematics equations or any idea what that is.
Your initial velocity vo = 41 m/s
You should ignore any wind considerations and the gravity constant you need is 9.8 m/s2
wadesweatt
Aug25-08, 11:54 PM
Ok thanks, I think I need this equation:
x = x0 + v0(t) + (1/2) a t^2
In order to find how high, I am looking for x, right? That stands for displacement I assume.
So, I would substitute to get x= 0 + 41(t) + (1/2)(9.8)(t^2). Original position (xO) should be zero, right? Since it starts at the ground... and what should time be?
wadesweatt
Aug26-08, 12:22 AM
....?
LowlyPion
Aug26-08, 01:54 AM
Ok thanks, I think I need this equation:
x = x0 + v0(t) + (1/2) a t^2
In order to find how high, I am looking for x, right? That stands for displacement I assume.
So, I would substitute to get x= 0 + 41(t) + (1/2)(9.8)(t^2). Original position (xO) should be zero, right? Since it starts at the ground... and what should time be?
You might figure the time a little differently first. You know that Vo is 41 m/s and you know that gravity will act to slow that down at 9.8 m/s2. The time to do that then is found with v=at or in this case you can solve for t with t = Vo/g were g is the 9.8 m/s2
Knowing the time then let's you find the distance with the simpler form of the equation x = 1/2 g * t2