91Ó°ÊÓ

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 \(PE(t)\): \(TE(t) = \frac{1}{2}m((v_{0}\cos{\theta_{0}})^2 + (v_{0}\sin{\theta_{0}} - gt)^2) + mg(v_{0}\sin{\theta_{0}}t - \frac{1}{2}gt^2)\)