Projectile Motion baseball hit at ground level

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a baseball hit at ground level. The scenario includes determining the height of a fence that the baseball barely clears after reaching its maximum height.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the time to maximum height and its implications for calculating the maximum height. There is mention of the relationship between the time to maximum height and the total time of flight in parabolic motion.

Discussion Status

Some participants have offered insights into the properties of projectile motion, particularly regarding the time to maximum height and its relevance to the overall motion of the baseball. There is an ongoing exploration of these concepts without a clear consensus on the next steps.

Contextual Notes

The original poster expresses uncertainty about how to approach the problem, indicating a potential gap in understanding the application of the equations of motion in this context.

rasikan
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Homework Statement



A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after begin hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 320 ft from where it was hit. How high is the fence?

Homework Equations


i think I can use these equation
x-x0=vx0t
y-y0=vyot-gt^2/2

The Attempt at a Solution


I have no clue about this question
 
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I would start with the fact that you know the time to maximum height. Since it is parabolic motion, you can use this to find the maximum height since you know that at the instant of maximum height, Vy=0. See if that helps.
 
Another consequence of parabolic motion is that the ball reaches its maximum height in half the time it takes to reach the ground again.
 
arunma said:
Another consequence of parabolic motion is that the ball reaches its maximum height in half the time it takes to reach the ground again.
This can also be proved by using
v-u/t=a
 
Last edited:

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