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

AI Thread Summary
The discussion revolves around calculating the maximum height an apple reaches when thrown at a speed of 14 m/s and an angle of 55 degrees. The initial height of 2.2 meters, where the friend's balcony is located, is deemed irrelevant to the maximum height calculation. The correct approach involves using the equations of motion to find the time of flight and then the maximum height without adding the balcony height. The final calculated maximum height is approximately 8.99 meters. Clarification on the problem's wording helped resolve confusion regarding the initial height value.
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+vot+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+vot+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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