A rugby player kicks the ball with an initial velocity of 40 m/s at an angle of 10 degrees. The goal posts are 40 m away and 2.2 m high. Will the ball go over the posts?

$$v_{0x} = v_0 \cos \theta = 40 \cos(10^\circ) \text{ m/s}$$ $$v_{0y} = v_0 \sin \theta = 40 \sin(10^\circ) \text{ m/s}$$

Using a calculator:

$$v_{0x} \approx 40 \times 0.9848 = 39.392 \text{ m/s}$$ $$v_{0y} \approx 40 \times 0.1736 = 6.944 \text{ m/s}$$

The horizontal motion is uniform (constant velocity).

$$x = v_{0x} t$$ $$40.0 \text{ m} = (39.392 \text{ m/s}) t$$ $$t = \frac{40.0}{39.392} \text{ s}$$ $$t \approx 1.0154 \text{ s}$$

The vertical motion is uniformly accelerated.

$$y = v_{0y} t - \frac{1}{2}gt^2$$ $$y = (6.944 \text{ m/s})(1.0154 \text{ s}) - \frac{1}{2}(9.8 \text{ m/s}^2)(1.0154 \text{ s})^2$$ $$y = 7.050 \text{ m} - (4.9 \text{ m/s}^2)(1.0310 \text{ s}^2)$$ $$y = 7.050 \text{ m} - 5.052 \text{ m}$$ $$y = 1.998 \text{ m}$$

The height of the ball when it reaches the goal post is $1.998 \text{ m}$.
The height of the goal post is $2.2 \text{ m}$.
Since $1.998 \text{ m} < 2.2 \text{ m}$, the ball passes under the crossbar. It is also above the ground ($0 \text{ m}$). Therefore, the ball hits the target (the goal area).

$$\boxed{\text{The ball hits the target.}}$$