# Need a little help

1. Sep 23, 2007

### jbutl3r25

Ok this is a follow-up type of question. The first problem was : A pelican flying along a horizontal path drops a fish from a height of 3.7 m. The fish travels 8.2 m horizontally before it hits the water below. What is the pelicans initial speed?

vx
dx 8.2
t

vfY
voY 0
aY -9.8
dY -3.7
t

I worked this out to eventually get to vx = 9.4, giving the pelican's speed.

Now the part that's giving me trouble is this:
If the pelican is traveling at the same speed, but only 3.3 m above the water, how far would the fish travel horizontally before hitting the water below? I used the same format that I did about, and switched the dY to -3.3 m...I got the answer and put that as my answer in the online homework service we use. It said the first part was correct, but this part is wrong. Anoybody see what I'm doing wrong?

2. Sep 23, 2007

### mgb_phys

You need to work out the time taken for the fish to freefall 3.3m using s = ut+1/2at^2
then put that time with the horizontal speed to get the horizontal distance.

3. Oct 7, 2007

### cat_eyes

help?

I have this same problem for my honors physics class and I was wondering what equation you used to get vx = 9.4?

4. Oct 7, 2007

### mgb_phys

The important point is that the fish's horizontal speed doesn't change - since no forces act horizontally.
So just use s = 1/2 g t^2 to get the freefall time and use that with the horizontal speed to get the horizontal distance travelled.