# Maximum height of the ball

1. May 15, 2014

### rlee1089

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
v=vo+at
y=yo+vot+1/2at^2

3. 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?

2. May 15, 2014

### 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.

Last edited: May 15, 2014
3. May 15, 2014

### goraemon

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.

4. May 15, 2014

### rlee1089

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