A heavy point is launched in the vertical plane Oxy

A heavy point is launched in the vertical plane $Oxy$ at an angle $\alpha$ to the horizon with initial speed $v_{0}$. Neglecting air resistance, find the equations of motion, the velocity, the acceleration and the trajectory.

Step 1: Resolve into components

\[ v_{0x} = v_{0}\cos\alpha, \qquad v_{0y} = v_{0}\sin\alpha. \]

Gravity acts downward only, so the two coordinate motions are independent: $\frac{d^{2}x}{dt^{2}} = 0, \qquad \frac{d^{2}y}{dt^{2}} = -g$.

Step 2: Integrate

\[ \frac{dx}{dt} = v_{0}\cos\alpha, \qquad \frac{dy}{dt} = v_{0}\sin\alpha-gt, \]

\[ x = v_{0}t\cos\alpha, \qquad y = v_{0}t\sin\alpha-\frac{gt^{2}}{2}. \tag{1} \]

The horizontal motion is uniform; the vertical one is Problem 343.

Step 3: Speed

\[ v^{2} = v_{0}^{2}\cos^{2}\alpha+\bigl(v_{0}\sin\alpha-gt\bigr)^{2} = v_{0}^{2}\bigl(\cos^{2}\alpha+\sin^{2}\alpha\bigr)-2v_{0}gt\sin\alpha+g^{2}t^{2}, \]

\[ v = \sqrt{v_{0}^{2}-2v_{0}gt\sin\alpha+g^{2}t^{2}}. \]

Step 4: Acceleration

\[ w_{x} = 0, \qquad w_{y} = -g \;\Longrightarrow\; w = g, \ \text{ constant and vertical}. \]

Step 5: Trajectory

Eliminating $t = \dfrac{x}{v_{0}\cos\alpha}$ from (1): $y = x\tan\alpha-\frac{g}{2v_{0}^{2}\cos^{2}\alpha}\,x^{2}$, a parabola with vertical axis, opening downward.

Step 6: Verification

\[t = 0:\ v = v_{0} \;\;\checkmark; \qquad \alpha = \frac{\pi}{2}:\ x \equiv 0 \text{ and the motion reduces to Problem 343} \;\;\checkmark\]

\[\text{Range: } y = 0 \text{ at } x = \frac{2v_{0}^{2}\sin\alpha\cos\alpha}{g} = \frac{v_{0}^{2}\sin2\alpha}{g}, \text{ greatest at } \alpha = 45^{\circ} \;\;\checkmark\]

\[\text{Summit: } t = \frac{v_{0}\sin\alpha}{g},\ y_{\max} = \frac{v_{0}^{2}\sin^{2}\alpha}{2g} \;\;\checkmark\]