Vertical throw

You throw a ball straight up from your hand. For a short moment it flies slower and slower, almost stops at the top, then falls back. That is a vertical throw: one-axis motion under constant gravitational acceleration.

At launch you give it speed v₀ upward. After that, in this model, only weight acts. We ignore air drag, so acceleration has constant magnitude g = 9.81 m/s² and a fixed downward direction. From that you can find maximum height, times, and speed at any instant.

Pick an axis. A convenient choice: y upward, y = 0 at launch level. Then a_y = -g, and for an upward throw v_y0 = +v₀. The uniformly accelerated equations become:

v_y(t) = v₀ - g·t

y(t) = v₀·t - ½·g·t²

Keep the units consistent: speed in m/s, time in seconds, height in meters, g in m/s². If your speed is in km/h, divide by 3.6. Example: 36 km/h is exactly 10 m/s.

At the apex the instantaneous speed drops to zero: v_y = 0. From the time-free relation v² = v₀² - 2·g·H you get the classic result:

H_max = v₀² / (2g)

Rise time follows from the same condition. From v₀ - g·t_up = 0 you get:

t_up = v₀ / g

First numeric intuition: with v₀ = 10 m/s and g = 9.81 m/s² you get H_max ≈ 5.10 m, t_up ≈ 1.02 s, and a full up-and-back flight to the same level of ≈ 2.04 s. The same numbers appear in the example below: work them out by hand, then check in the vertical-throw calculator.

Second intuition: double v₀ and H_max grows fourfold, because the formula has v₀². With v₀ = 20 m/s and the same g you get H_max ≈ 20.4 m, not “about ten meters”. Rise time only doubles (t_up ≈ 2.04 s), so height grows faster than time.

If launch and landing share the same height, the up and down legs are symmetric. Total flight time is T = 2·t_up = 2·v₀/g. On return the speed matches the launch value, only the direction is down. That follows from constant g and no drag, not from anything special about the ball.

The state at time t comes straight from the equations: height y(t) and speed v(t). After the apex v is negative (with an upward axis) while height can still be positive. In a problem, watch the signs or describe “up / down” in words so the direction does not get lost.

The same H_max shows up in energy. With no drag, Em = Ek + Ep is constant. At launch (y = 0): Em = ½·m·v₀². At the apex: Em = m·g·H_max. Hence, again H_max = v₀²/(2g). Energy checks height well, but it does not give you the time on its own. Compute launch Ek in kinetic energy and apex Ep in potential energy.

Free fall is the vertical-throw limit with v₀ = 0: no rise phase, only downward distance h = ½·g·t² and v = g·t. If the problem says “drop”, do not invent an upward H_max. If it says “throw downward at v₀”, the launch component aligns with g and distance grows faster than for a pure drop. For a drop, use free fall.

If you throw upward from a balcony and land on the ground below the start, classic T = 2·v₀/g no longer holds. Then solve the position equation for the landing level, with an explicit h₀. Same formalism, different end condition.

A common mistake sounds like this: “at the top the speed is zero, so acceleration must be zero too.” It feels convincing, because the ball briefly “hangs”. But instantaneous speed is not the cause of the motion. g still acts downward, so right after the apex the body speeds up again. Velocity is zero; acceleration is not.

Other traps: thinking mass affects H_max (with no drag, m cancels), or confusing H_max with horizontal range when the problem has no x axis at all. Without drag, up and down times to the same level match; the feeling that “down is faster” usually comes from drag or from a different landing level.

Think of it this way: g removes about 9.81 m/s² of upward speed each second. After time v₀/g the upward budget is gone. On the way down the same g rebuilds the speed. Until wind and drag appear, the story is symmetric.

On a graph, v(t) is a straight line of slope -g, and y(t) is a parabola. The apex is where the v(t) line crosses zero. The area under v(t) on the way up is H_max. That geometry helps when you forget the algebraic formula but remember the shape of uniformly accelerated motion.

The model breaks down with strong drag, a light object at large v₀, or when landing is clearly below launch level and you still force T = 2·v₀/g. Then the H_max formula is more of an upper estimate. For everyday ball problems at modest v₀, constant g is good enough.

Before you calculate, check in order: (1) whether landing is at launch level, (2) whether v₀ is up, down, or zero, (3) whether the problem uses g = 9.81 m/s² or 10, (4) whether the units match the choice in the page header, (5) whether the answer looks sensible (at 10 m/s you expect a few meters, not kilometers). Then check the result in the calculator.

When a horizontal component appears, the vertical part of the flight still obeys the same equations. A vertical throw is therefore not only a standalone problem, but also the “y module” inside projectile motion. That is why H_max and t_up belong in working memory. The next topic in the series is the horizontal throw.