What Initial Speed Does the Second Kingfisher Need to Catch the Dropped Fish?

[LastBloomingFlower]

So I can NOT understand what to even do first... Please help me!

A kingfisher is 30m above a lake when it accidentally drops a fish. A second kingfish 5m above the first dives toward the fish,. What's the initial speed it needs to catch the fish just before it reaches the water?

 

[BishopUser]

I assume that the 2nd bird will go into a nose dive and its acceleration will simply be gravity.

first realize that the initial velocity of the fish will be 0 since it is simply dropped so start with the projectile motion equation...

xf = xi + vit + 1/2at^2

xf = final position
xi = initial position
vi = intial velocity
t = time
a = acceleration

set the 0 level at the 2nd bird...

xf = 35m
xi = 5m
vi = 0
t = unknown
a = 9.8m/s^2

plug those into the equation to get time = 2.47s
so now we know that the 2nd bird has 2.47 seconds to catch the fish before it hits the water

use the same equation with the info for the 2nd bird.

xf = xi + vit + 1/2at^2
35 = 0 + (vi)(2.47) + (1/2)(9.8)(2.47)^2
vi = 2.08 m/s

so the 2nd bird had to start his dive going 2.08m/s to be able to catch the fish.

 

[LastBloomingFlower]

Ok Thanks so much. I started out that way but I got stuck!