How High Does the Apple Reach When Thrown at 55 Degrees?

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically the maximum height reached by an apple thrown at an angle of 55 degrees with an initial speed of 14 m/s. The apple is thrown towards a balcony that is 2.2 meters above the release point.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum height using kinematic equations but questions the relevance of the 2.2m height of the balcony in their calculations. Other participants discuss the potential misunderstanding of this information.

Discussion Status

Participants are exploring the implications of the balcony height on the problem. Some suggest that the 2.2m height may not be necessary for determining the maximum height of the apple, while others express uncertainty about the professor's comments regarding the initial conditions.

Contextual Notes

There is a mention of a previous quiz context, and the discussion reflects on the interpretation of the problem statement and the relevance of certain details provided in the problem.

rlee1089
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Homework Statement


You want to throw an apple to your buddy on the second floor balcony that is 2.2m above where you release the apple. When you throw the apple with a speed of 14m/s at an angle of 55° above the horizontal, it gets to your buddy who catches it on the way down. Find the maximum height of the ball.


Homework Equations


v=vo+at
y=yo+volt+1/2at^2

The Attempt at a Solution


I solved for [t] using v=vo+at in the y direction:
0=11.47-9.8t
t=1.17s

I plugged in the [t] to solve for [y], maximum height:
y=2.2+11.5(1.17)+1/2(-9.8)(1.17)^2
y=8.99m

This was a quiz from weeks ago. My professor scribbled something about how I shouldn't "add that" in the above equation, which I'm suspecting is the part about 2.2m. If that's the issue, why shouldn't I put 2.2m as my initial y value?
 
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Your answer seems right to me (approximately... I got 8.91, but you approximated before the end so that can justify the difference).

I don't know what your professor meant, perhaps you should ask him/her.

Edit: Good call, goraemon; Lesson learned: Read carefully.
 
Last edited:
rlee1089 said:

Homework Statement


You want to throw an apple to your buddy on the second floor balcony that is 2.2m above where you release the apple. When you throw the apple with a speed of 14m/s at an angle of 55° above the horizontal, it gets to your buddy who catches it on the way down. Find the maximum height of the ball.


Homework Equations


v=vo+at
y=yo+volt+1/2at^2

The Attempt at a Solution


I solved for [t] using v=vo+at in the y direction:
0=11.47-9.8t
t=1.17s

I plugged in the [t] to solve for [y], maximum height:
y=2.2+11.5(1.17)+1/2(-9.8)(1.17)^2
y=8.99m

This was a quiz from weeks ago. My professor scribbled something about how I shouldn't "add that" in the above equation, which I'm suspecting is the part about 2.2m. If that's the issue, why shouldn't I put 2.2m as my initial y value?

The problem states that the friend's balcony is 2.2m above where you release the apple. So that's why you shouldn't put 2.2m as your "initial y value."

In fact, the whole bit of information surrounding 2.2m is a red herring - it is completely irrelevant to the answer you're asked to find - which is simply, how high does the ball go given that you released it with an initial speed of 14m/s at 55° above horizontal.
 
goraemon said:
The problem states that the friend's balcony is 2.2m above where you release the apple. So that's why you shouldn't put 2.2m as your "initial y value."

In fact, the whole bit of information surrounding 2.2m is a red herring - it is completely irrelevant to the answer you're asked to find - which is simply, how high does the ball go given that you released it with an initial speed of 14m/s at 55° above horizontal.

I see it now...So it was just an irrelevant piece of information, eh? Thank you for clarifying that! It was driving me insane. :P
 

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