A bullet of mass 25 grams and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5kg suspended from a chord 3 m long.

How far will the block swing from its position of rest before beginning to return.

so, the bullet has a momentum = mv = 0.025 x 200 = 5 kg m/s

momentum after embedding in the block = (5 + 0.025).u

conservation of momentum gives u = 5/5.025 = 0.995 m/s

so, the pendulum starts to swing at u = 0.995 m/s and will be retarded by the weight of the bullet/block combination. Now according to my book, the angle of the pendulum is calculated by calculating the vertical height of the pendulum as it swings upward to a stop. i.e.

$ v^2 = 2gh $ ............................(1)

where $g = 10m/s^2$ and $h = 3(1 - \cos \theta)$ where $\theta$ is the angle from the vertical.

$ 0.99 = 2. 10. 3(1 - \cos \theta) $

This gives the right answer, but I don't see how it works.

The velocity is initially horizontal, the formula (1) above applies in the vertical direction, where the initial velocity is zero.

Also the equations of motion refer to motion in a straight line, can they be applied to circular motion, or are we just making an approximation?