A Projectile Is Horizontal Constant Velocity Plus Vertical Free Fall
Start two balls from the same height at the same time. One you simply drop; the other you throw sideways. Drag time and something striking shows up: the ball thrown sideways passes through exactly the same height as the dropped one, and hits the ground at the same moment. However fast you throw it horizontally, it falls by the same amount. Here is the secret: projectile motion is really two separate motions overlaid. Horizontally there is no force, so constant velocity; vertically there is only gravity, so constant acceleration. Because the two directions do not interfere and are independent, you apply the formulas you already learned to each one and simply combine them at the end. A complicated motion that looks like a curve turns out to be a combination of two straight lines.
Drag time t to move both balls at once. The left ball was simply dropped; the right ball was thrown horizontally from the same height. The dashed line joins their heights, and at every instant the two heights are exactly equal. No matter how far the thrown ball flies sideways, it falls by the same amount as the one merely dropped, so the two hit the ground at the same moment. This means the horizontal motion does not interfere with the vertical fall at all. Gravity does not care how fast you go sideways; it only pulls downward. This independence is the key to solving projectile motion.
Now take just the horizontal direction. Horizontally there is no force pushing or resisting the ball (ignoring air resistance). With no force the acceleration is zero, so the horizontal velocity never changes from start to finish: constant-velocity motion. Drag the vx slider and the marks, dropped at equal time intervals, always sit equally far apart. Throw faster and the gaps widen, slower and they narrow, but in every case the spacing is uniform. So the horizontal position is simply x = vx·t, the simplest motion of all, in direct proportion to time. The horizontal axis of the parabola runs at the same speed the whole way through.
Now look at just the vertical direction. There is an initial upward speed, but gravity keeps pulling down, so the vertical motion is exactly the constant-acceleration motion of the last lesson, with acceleration always -g. Drag time t and the ball rises, slows, pauses for an instant at the top, then comes back down, with the way up and the way down mirror-symmetric. The height is y = vy·t - ½g t², precisely the constant-acceleration formula. The top is the moment the vertical velocity reaches zero, where it reaches its maximum height. Forget the horizontal; looking only up and down, it is just the same as throwing a ball straight up.
Now combine the two motions. At every instant the horizontal position advances steadily to the right as x = vx·t, while the vertical position rises and falls as y = vy·t - ½g t². Drag time t and watch: horizontally it steps along at an even pace, vertically it traces an arc, and the point where the two meet flies along a smooth curve. That curve is the parabola. The gray scale below shows the horizontal progress and the one at the side shows the vertical height; the ball's position is always where those two shadows meet. It looks like a curve, but nothing here is new: it is just horizontal constant velocity and vertical constant acceleration, the two motions learned separately, run at once on the same clock.
Finally, change the launch angle. Keep the launch speed fixed and turn only the angle with the slider, and even with the same effort, how far and how high it goes changes sharply. Raise the angle and it climbs higher but does not reach as far; lower it and it stays flat and again falls short. The range is longest at the exact middle, 45 degrees. And there is a neat symmetry: two angles equally far from 45, such as 30 and 60, give the same range. One simply flies high and lazy, the other low and fast. The maximum height keeps growing as the angle steepens, peaking at 90 degrees, straight up. The range is given by R = v0² sin2θg.