Projectile Motion: Calculating Time to Shoot a Flying Bird

[fishingspree2]

Homework Statement

A bird is flying horizontally at a constant velocity of 2 m/s. 10 meters under the bird, there is a man that wants to shoot the bird. He is holding his gun at an angle of 45 degrees from horizontal. The initial velocity of the bullet is 25 m/s. When the man must open fire?

Homework Equations

x = x.init + x.velocity.init * time + 0.5 * accel.x * time^2
y = y.init + y.velocity.init * time + 0.5 * accel.y * time^2
time=t

The Attempt at a Solution

First, I find the x and y components of the initial velocity of the bullet:
x.velocity.init.bullet = 25 * cos(45 degrees) = 17.68 m/s
y.velocity.init.bullet = 25* sin(45 degrees) = 17.68 m/s

Then, since the bird always stays at a height of 10 meters, I find how much time the bullet takes to reach this height.
y = y.init + y.velocity.init * time + 0.5 * accel.y * time^2
10 = 0 + 17.68*t + 0.5*-9.8*t^2
t = 0.702 s

I find the horizontal position of the bullet after 0.702 seconds
x = 17.68m/s * 0.702 s
x = 12.41 m

Now, I find how much time the bird takes to fly 12.41 meters
x = initial x + x.velocity.bird * time
12.41 = 0 + 2*t
t = 6.205 s

The final answer is 6.205-0.702 = 5.503 seconds

Is my work correct?
Sorry for my bad english.

 

[gabbagabbahey]

It looks pretty good to me: I get t= 5.51s which means that the hunter must shoot 5.51s after the bird is over top of him, or in terms of distance, when the bird is 11.01m away from him.

 

[Mattowander]

Work looks alright to me, but I just skimmed over it quickly.