# Homework Help: Projectile + Conservation of energy

1. Oct 23, 2005

### Neerolyte

I have no idea how i keep getting the wrong answer. Here's what i did
At the top of the trajectory the horizontal velocity is the same which is
cos(35)(125).

By using conservation of momentum i can find the velocity of fragment 2

MVix = MVfx
(28)(cos(35)(125)) = (14)(0) + (14)(Vf2x)
So i just find Vf2 which is 204.788011 m/s

Well to calculate range i need to know the max height, but we also need the time it reaches max height. So to find time that reaches max height:

Vfy = Voy + ayt
0 = sin(35)(125) + (-9.81)t
t = 7.3085s

Max Height:
dy = Voyt + 0.5ayt
dy = (sin(35)(125))x(7.3085) + 0.5(-9.81)(7.3085)
dy = 262m

We now know the max height, horizontal velocity of fragment 2, isn't this like the simple projectile motion question where:

(Horizontal velocity of fragment 2) / (time it takes for the object to fall 262m) = the range of fragment 2

then you add that up to the range it will take the projectile to get to max height.

If it's not clear enough please tell me so i can rephrase it or something.

2. Oct 23, 2005

### Fermat

Thanks for posting your working. It helps a lot.

Everything you have done, except one part, is correct and quite clear.

(Horizontal velocity of fragment 2) / (time it takes for the object to fall 262m) = the range of fragment 2

That should be a multiply sign, not a division sign, giving

(Horizontal velocity of fragment 2) x (time it takes for the object to fall 262m) = the range of fragment 2

There's just a semantic problem, perhaps.
The Fragment2 doesn't actually exist until until the explosion a the top of the flight. So, its range may simply be the horizontal distance travelled in the time it takes to fall to the ground.

Last edited: Oct 23, 2005
3. Oct 23, 2005

### Neerolyte

Just calulating the range of Fragment 2, i still got the wrong answer. (by the way it's an electronic questionaire that tells you if you get the answer right or wrong if you submit the answer).

And i checked all the numbers many many times.

I don't know if this is allowed but could somebody do the question and can we compare answers?

and the Range of the Fragement is roughly 28m

4. Oct 23, 2005

### Fermat

I made a mistake in my first post. I'm afraid I mis-read your text. I really should wear my specs.

(Horizontal velocity of fragment 2) / (time it takes for the object to fall 262m) = the range of fragment 2

That should be a multiply sign, not a division sign, giving

(Horizontal velocity of fragment 2) x (time it takes for the object to fall 262m) = the range of fragment 2

The time to fall the 262 m is the same as the time to get up there. i,e t = 7.3085 s.
Horizontal velocity is Vx = 204.788011 m/s
Range = (Horizontal velocity) x (The time to fall the 262 m) = 204.788011 m/s x 7.3085 s. = 1496.693 m

Range = 1496.693 m
===============

Last edited: Oct 23, 2005
5. Oct 23, 2005

### Neerolyte

OH my goodness!!! Thank you so much!!!
and by the way the wording of the question is SOOO bad, because when i enter 1496.693 it says i'm wrong, because the question is refering to the range from the initial firing position. While the question is misleading to us thinking that it's just the range of Fragment 2, but i got the right answer. Thanks a lot