Two dimensional motion problem.

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Homework Help Overview

The problem involves a baseball hit at a speed of 45 m/s at a 45-degree angle, focusing on determining the time it remains in the air and the horizontal distance it travels. This falls under the subject area of projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations relevant to projectile motion, including the calculations for horizontal and vertical components of velocity. There are inquiries about the correctness of the derived formulas for time and range.

Discussion Status

Some participants have provided equations and calculations related to the problem, while others express uncertainty about the correctness of these approaches. There is an acknowledgment of the equations being used, but no consensus on their application has been reached.

Contextual Notes

The original poster requests hints or formulas, indicating a need for guidance without direct solutions. The discussion reflects a collaborative effort to clarify the problem setup and relevant equations.

TheSilence
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Homework Statement


A baseball was hit at 45m/s at an angle of 45 degrees above the horizontal.
How long did it remain in the air? How far did it travel horizontally?

Hi everyone, I am i stuck :cry: on this problem any hints (formulas) to help me solve this problem would be appreciated. Thanks in advance
 
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Today seems to be the projectile motion day. :smile:

Anyway, which are the equations for projectile motion? Start with finding the equations for the x and y displacement of the baseball.
 
So I need to find the time and range of the projectile.
[tex]V_0x[/tex]= 45m/s cos(45)
[tex]V_0y[/tex]=45m/s sin(45)

and Range is: [tex]R = 2{V_0x}{V_oy}/{g}[/tex]
Time is: [tex]t = {2V_0y}/{g}[/tex]

:confused: am I wrong?
 
TheSilence said:
So I need to find the time and range of the projectile.
[tex]V_0x[/tex]= 45m/s cos(45)
[tex]V_0y[/tex]=45m/s sin(45)

and Range is: [tex]R = 2{V_0x}{V_oy}/{g}[/tex]
Time is: [tex]t = {2V_0y}/{g}[/tex]

:confused: am I wrong?

Good work, you're right. :wink:
 
Thanks radou :approve: !
 

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