A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) at angle \(\theta_{0}\) above the horizontal. Air resistance is negligible. a) Calculate the angular momentum of the projectile about the origin. b) Calculate the rate of change of this angular momentum. c) Calculate the torque acting on the projectile, about the origin, during its flight.

The projectile motion can be divided into horizontal and vertical components. We can use the given values and write the equations for horizontal and vertical motions: Horizontal motion: $$ x(t) = v_{0x}t \\ v_{0x} = v_0 cos(\theta_0) $$ Vertical motion: $$ y(t) = v_{0y}t - \frac{1}{2}gt^2 \\ v_{0y} = v_0 sin(\theta_0) $$ Where \(t\) is the time, \(g\) is the acceleration due to gravity, \(v_{0x}\) and \(v_{0y}\) are the horizontal and vertical components of the initial velocity, respectively.

The position vector \(r\) and velocity vector \(v\) can be written as: $$ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} \\ \vec{r}(t) = (v_0 cos(\theta_0) t) \hat{i} + (v_0 sin(\theta_0) t - \frac{1}{2}gt^2)\hat{j} $$ $$ \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = v_{0x}\hat{i} + (v_{0y} - gt)\hat{j} \\ \vec{v}(t) = v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j} $$

Angular momentum \(L\) with respect to the origin is given by: $$ \vec{L} = \vec{r} \times m\vec{v} $$ Using the cross product, we can find the angular momentum: $$ \vec{L} = (v_0 cos(\theta_0) t) \hat{i} \times m(v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j}) + (v_0 sin(\theta_0) t - \frac{1}{2}gt^2)\hat{j} \times m(v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j}) $$ The \(\hat{k}\) component of the angular momentum is: $$ L_z = m(v_0 cos(\theta_0) t)(v_0 sin(\theta_0) - gt) - m(v_0 sin(\theta_0) t - \frac{1}{2}gt^2)(v_0 cos(\theta_0)) $$ Simplifying this expression, we get: $$ L_z = m(v_0^2t sin(\theta_0) cos(\theta_0) - \frac{1}{2}gt^2 v_0 cos(\theta_0)) $$ The angular momentum of the projectile about the origin is \(L_z\).

To find the rate of change of angular momentum, we need to find the derivative of \(L_z\) with respect to time: $$ \frac{dL_z}{dt} = m(v_0^2 sin(\theta_0)cos(\theta_0) - gt v_0 cos(\theta_0)) $$

As there's no external torque acting on the projectile during its motion (negligible air resistance), according to Newton's laws, the rate of change of angular momentum should be equal to the torque \(\vec{\tau}\) acting on the projectile: $$ \vec{\tau} = \frac{dL_z}{dt} $$ So, the torque acting on the projectile during its flight is \(\vec{\tau} = m(v_0^2 sin(\theta_0)cos(\theta_0) - gt v_0 cos(\theta_0))\hat{k}\). In this exercise, we have calculated the angular momentum of a projectile, its rate of change, and the torque acting on it during its flight, using kinematic equations, the definition of angular momentum, and Newton's laws.