I've attached a picture which include the question -- I found it on the net, and it's exactly the same as my homework.

I've got hints on this before, but I don't understand how to calculate the length, so please show me how if you can

2. Originally Posted by KiyoEtAlice
I've attached a picture which include the question -- I found it on the net, and it's exactly the same as my homework.

I've got hints on this before, but I don't understand how to calculate the length, so please show me how if you can
Actually you are dealing with a right triangle. (see attachment)

According to Euklid's theorem you have:

$\left(\dfrac d2 \right)^2 = h \cdot (2R-h)$

Solve for R.

3. thanks but can you tell me what "d" value is supposed to be? I think I got things mized up. "h" is 5 for the ball right?

4. Originally Posted by KiyoEtAlice
thanks but can you tell me what "d" value is supposed to be? I think I got things mized up. "h" is 5 for the ball right?
The text of the problem says: "... They both travel 10 m in the x-direction. ..." So here you have d = 10 m

5. Originally Posted by earboth
The text of the problem says: "... They both travel 10 m in the x-direction. ..." So here you have d = 10 m

Okay Thanks!

I've done it, and my R for each of the cases come out differently, is this supposed to happen?

6. Originally Posted by KiyoEtAlice
Okay Thanks!

I've done it, and my R for each of the cases come out differently, is this supposed to happen?
of course. The curve of the ball is much more bent than the curve of the bullet. Thus the radius connected to the curve of the ball is much smaller than the radius of the curve of the bullet.

I've got $r_{ball} = 5\ m$ and $r_{bullet} = 25000\ m$

7. Originally Posted by earboth
of course. The curve of the ball is much more bent than the curve of the bullet. Thus the radius connected to the curve of the ball is much smaller than the radius of the curve of the bullet.

I've got $r_{ball} = 5\ m$ and $r_{bullet} = 25000\ m$
Okay, thanks! I've got the same result. Thanks a lot!