Confusion about Instructor's solution of HRK

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Homework Help Overview

The discussion revolves around the interpretation of an instructor's solution related to projectile motion, specifically concerning the initial speed of a ball and its relation to the range and height of the projectile.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of the instructor's solution, particularly the assumption that the final position of the projectile is at the same height as the initial position. There is a discussion about the implications of this assumption on the accuracy of the initial speed calculation.

Discussion Status

Some participants express agreement with the concerns raised about the instructor's solution, noting potential errors in the solution manual. There is an exploration of different interpretations regarding the definition of horizontal range and its implications for the problem.

Contextual Notes

Participants mention that the solution manual may have misinterpreted the horizontal range as the distance traveled before returning to the initial height, which could lead to confusion in the context of the problem.

phymath7
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Homework Statement
This is an excercise from the book 'Physics' by HRK:A batter hits a pitched ball at a height 4.0 ft above the ground so that its angle of projection is 45° and its horizontal range is 350 ft. The ball travels down the left field line where a 24-ft high fence is located 320 ft from home plate. Will the ball clear the fence? If so, by how much?
Relevant Equations
##y-y_0=xtan\theta-\frac {gx^2}{2v_0^2cos^2\theta}##
The instructor's solution goes like this:
The initial speed of the ball is given by ##v_0=\sqrt{gR}## where R is the range.But this is true if the final position of the projectile is y=0 but in this case, y=-4.Though this doesn't affect much in this case,but for higher velocity and extreme cases this certainly would.Am I right?
 
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Yes, you are right.
 
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TSny said:
Yes, you are right.
Then it's a pity that this solution manual contains such basic error.😑
 
phymath7 said:
Then it's a pity that this solution manual contains such basic error.😑

Perhaps the solution in the manual interprets the horizontal range R (350 ft) to be the horizontal distance that the ball would travel before it returns to its initial height of 4 ft above the ground (assuming the ball clears the fence). Then, ##v_0 = \sqrt{gR}## is OK. I'm not defending this interpretation, but it might be what was going on in the mind of whoever wrote the solution. Who knows. I'm with you in interpreting the horizontal range as the horizontal distance traveled until landing on the ground.

It's fairly common for solution manuals to have occasional errors.
 
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