Conservation of Energy

ES | EN

We consider a projectile launched at an angle with constant gravity \(g\) (vertical) and a possible external horizontal acceleration \(a_x\). If \(v_0\) is the initial speed, \(\theta\) the launch angle, and \(y_0\) the initial height, then:

\[ \begin{aligned} x(t) &= v_0\cos\theta \; t \;+\; \tfrac{1}{2} a_x t^2,\\ y(t) &= y_0 \;+\; v_0\sin\theta \; t \;-\; \tfrac{1}{2} g t^2,\\[4pt] v_x(t) &= v_0\cos\theta \;+\; a_x t,\qquad v_y(t) = v_0\sin\theta \;-\; g t. \end{aligned} \]

The instantaneous energies are:

\[ E_c(t)=\tfrac{1}{2}m\!\left(v_x^2+v_y^2\right),\qquad E_p(t)=mgy,\qquad E_m(t)=E_c(t)+E_p(t). \]

Time (t): 0.00 s

Position (x, y): (0.00, 0.00) m

Velocity (vₓ, vᵧ): (0.00, 0.00) m/s

Kinetic Energy (E_c): 0.00 J

Potential Energy (E_p): 0.00 J

Total Mechanical Energy (E_m): 0.00 J