Throw something, and once it leaves your hand only one force acts on it: gravity, straight down.

Two motions, one clock#

Gravity pulls straight down, so it cannot change the horizontal motion, and no force of the throw carries the ball along: the hand stopped pushing at launch. What keeps it moving forward is inertia. Split the launch velocity into components and treat them as two one-dimensional problems that share a clock:

$$\vel{v_{x0}} = \vel{v_0}\cos\theta, \qquad \vel{v_{y0}} = \vel{v_0}\sin\theta$$

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The time comes from the vertical#

Horizontally there is no force, so $x = v_{x0}t$. Vertically there is a constant acceleration $-g$, the one-dimensional problem from the first lesson:

$$y = y_0 + v_{y0}t - \tfrac{1}{2}gt^2$$

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Where 45° stops being best#

On level ground with no air, the range is $R = v_0^2 \sin 2\theta / g$, so 45° throws farthest. Both conditions matter.

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What comes next#

You split a throw into two motions that share a clock, and saw 45° stop being best once height or air joined in. Next we ask what forces actually do, and test the intuition that a moving object needs something pushing it.