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

[rlee1089]

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?

 

[Nathanael]

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.

 

[goraemon]

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.

 

[rlee1089]

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

 

[HalfLostAndSearching]

The reason why you should not add 2.2m as your initial y value is because it is already included in the initial vertical velocity (vo) of the apple. In this problem, the initial vertical velocity of the apple is not 0m/s, it is 11.47m/s (found using the given speed and angle). Therefore, when you plug in the value of [t] in the equation for [y], you should not add 2.2m as this will result in double counting the initial vertical displacement. This is why your professor scribbled not to "add that" in the above equation. The correct equation for finding the maximum height of the ball would be:

y=11.47(1.17)+1/2(-9.8)(1.17)^2
y=8.99m

This will give you the correct answer of 8.99m, which is the maximum height of the ball.