A projectile of mass \(m\) is launched from the ground at \(t=0\) with a speed \(v_{0}\) and at an angle \(\theta_{0}\) above the horizontal. Assuming that air resistance is negligible, write the kinetic, potential, and total energies of the projectile as explicit functions of time.

First, we need to calculate the position of the projectile as a function of time. To do this, we can break down the position into horizontal (x-axis) and vertical (y-axis) components. For the horizontal component, the velocity remains constant (there's no horizontal acceleration) so the position can be given by: \(x(t) = v_{0x}t\) And for the vertical component, we'll use the equation of motion for constant acceleration: \(y(t) = v_{0y}t -\frac{1}{2}gt^2\) Since the projectile is launched at an angle \(\theta_0\) above horizontal, using trigonometry, we can write the horizontal and vertical initial velocities as follows: \(v_{0x} = v_{0}\cos{\theta_{0}}\) \(v_{0y} = v_{0}\sin{\theta_{0}}\) The position as a function of time can be given as \({x(t), y(t)}\): \(x(t) = v_{0}\cos{\theta_{0}}t\) \(y(t) = v_{0}\sin{\theta_{0}}t - \frac{1}{2}gt^2\)

Next, we need to calculate the velocity as a function of time. To do this, we'll take the derivative of the position functions found above with respect to time. \(v_x(t) = \frac{dx(t)}{dt} = v_{0}\cos{\theta_{0}}\) \(v_y(t) = \frac{dy(t)}{dt} = v_{0}\sin{\theta_{0}} - gt\) The velocity as a function of time can be given as \({v_x(t), v_y(t)}\).

Now, we can calculate the kinetic energy (KE) as a function of time. KE is given by the equation: \(KE = \frac{1}{2}m(v^2)\) Since we have the velocity as \({v_x(t), v_y(t)},\) we can find the square of the magnitude of the velocity as follows: \(v^2 = v_x(t)^2 + v_y(t)^2\) Plug in our expressions for \(v_x(t)\) and \(v_y(t)\): \(v^2 = (v_{0}\cos{\theta_{0}})^2 + (v_{0}\sin{\theta_{0}} - gt)^2\) Now, we can plug our expression for \(v^2\) into the KE equation: \(KE(t) = \frac{1}{2}m((v_{0}\cos{\theta_{0}})^2 + (v_{0}\sin{\theta_{0}} - gt)^2)\)

The potential energy (PE) is given by: \(PE = mgh\) We calculated the height, \(h,\) as a function of time, which is \(y(t)\): \(h(t) = v_{0}\sin{\theta_{0}}t - \frac{1}{2}gt^2\) Now, we can plug our expression for \(h(t)\) into the PE equation: \(PE(t) = mg(v_{0}\sin{\theta_{0}}t - \frac{1}{2}gt^2)\)

The total energy (TE) is the sum of the kinetic and potential energies: \(TE(t) = KE(t) + PE(t)\) Now, we can plug in our expressions for \(KE(t)\) and \(PE(t)\): \(TE(t) = \frac{1}{2}m((v_{0}\cos{\theta_{0}})^2 + (v_{0}\sin{\theta_{0}} - gt)^2) + mg(v_{0}\sin{\theta_{0}}t - \frac{1}{2}gt^2)\)