A.1.8—Projectile components

Separate the axes

With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:

$$u_x=u\cos\theta,\qquad u_y=u\sin\theta$$

Horizontal motion

There is no horizontal acceleration in the ideal model, so $v_x=u_x$ and $x=u_xt$. Use the horizontal displacement to find time or horizontal speed.

Vertical motion

Use one-dimensional constant-acceleration equations vertically, usually with $a_y=-g$ if upward is positive. The horizontal and vertical equations share the same time $t$.

Worked example from local Question Bank row 31723

A tennis ball travels $11.9\,\mathrm{m}$ horizontally after launch at $64.0\,\mathrm{m\,s^{-1}}$ and $7^\circ$ to the horizontal.

$$u_x=64.0\cos7^\circ=63.52\,\mathrm{m\,s^{-1}}$$
$$t=\frac{x}{u_x}=\frac{11.9}{63.52}=0.187\,\mathrm{s}$$

The same $0.187\,\mathrm{s}$ must then be used in the vertical equation.

Common trap

Do not use the launch speed as the horizontal speed. Resolve it first, and do not use the horizontal time independently of the vertical motion.