平抛运动性质¶

基础¶

$\Delta v = gt$

水平:

$\left\{ \begin{aligned} v_x &= v_0 \\ x &= v_0 t \end{aligned} \right.$

竖直:

$\left\{ \begin{aligned} v_y &= gt \\ y &= \frac{1}{2} g t^2 \\ \end{aligned} \right.$

合成:

$\left\{ \begin{aligned} v &= \sqrt{v_x^2 + v_y^2} \\ x &= \sqrt{x^2 + y^2} \\ \end{aligned} \right.$

推论¶

$\because \left\{ \begin{aligned} \tan{\theta} &= \dfrac{v_y}{v_x} = \dfrac{2y_A}{x_A} \\ \tan{\alpha} &= \dfrac{y_A}{x_A} \end{aligned} \right.$

$\therefore \tan{\theta} = 2 \tan{\alpha}$

$\tan{\theta} = 2 \tan{\alpha}$

$x_b = \frac{1}{2} x_a$

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