How High and How Long Does a Baseball Travel When Hit Straight Up at 41 m/s?

[wadesweatt]

Ball is hit up into the air...

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]

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]

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]

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]

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]

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.

This post addresses the equations you will need:

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Your initial velocity v_o = 41 m/s
You should ignore any wind considerations and the gravity constant you need is 9.8 m/s²

 

[wadesweatt]

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]

...?

 

[LowlyPion]

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 V_o 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 = V_o/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