# Nakafa Learning Content

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URL: https://nakafa.com/en/subjects/physics/kinematics/vertical-movement
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/physics/kinematics/vertical-movement/en.mdx

Learn vertical motion as one-dimensional uniformly accelerated motion with gravity always directed downward.

---

## A Vertical Axis Needs a Positive Direction

Vertical motion is upward or downward motion along one straight line. To keep the signs clear, we choose upward as positive and downward as negative.

With that choice, gravitational acceleration is written as $$-g$$ because gravity always pulls the object downward.

Visible text: With that choice, gravitational acceleration is written as because gravity always pulls the object downward.

Component: MathContainer
Children:

```math
y=y_0+v_0t-\frac{1}{2}gt^2
```

```math
v_y=v_0-gt
```

```math
v_y^2=v_0^2-2g(y-y_0)
```

Component: VerticalMovementLab
Props:
- title: Ball in Vertical Motion
- description: Choose how the ball starts moving to see how gravity changes its height
and velocity.
- labels: {
condition: "Initial condition",
modeLabels: {
throw: <>Thrown Upward</>,
drop: <>Dropped</>,
},
initialCondition: <>Initial condition</>,
maxHeight: <>Highest point</>,
time: <>Motion time</>,
finalVelocity: <>Final velocity</>,
viewLabel: "Ball in vertical motion visual",
}

## Gravity Still Acts While the Object Rises

When an object is thrown upward, its initial velocity is positive. After that, gravity reduces its vertical velocity little by little until the object stops for an instant at the highest point.

> At the highest point, the vertical velocity is momentarily zero, but gravitational acceleration still points downward.

This idea matters because students often think gravity disappears at the top. Only the velocity is zero for an instant, not the acceleration.

## An Upward Throw Stops for an Instant

Suppose a ball is thrown upward with initial velocity $$20\text{ m/s}$$ from position $$y_0=0$$. Use $$g=10\text{ m/s}^2$$. At the highest point, its vertical velocity is $$v_y=0$$.

Visible text: Suppose a ball is thrown upward with initial velocity from position . Use . At the highest point, its vertical velocity is .

Component: MathContainer
Children:

```math
v_y=v_0-gt
```

```math
0=20-10t
```

```math
t=2\text{ s}
```

The maximum height above the launch point can be found from the position equation.

```math
\begin{aligned}
y-y_0&=v_0t-\frac{1}{2}gt^2\\
&=20(2)-\frac{1}{2}(10)(2^2)\\
&=20\text{ m}
\end{aligned}
```

So the ball takes $$2\text{ s}$$ to reach the top and rises $$20\text{ m}$$ above its starting point.

Visible text: So the ball takes to reach the top and rises above its starting point.

## Reading the Sign of Velocity

The sign of $$v_y$$ tells the direction of motion at that moment. A positive value means the object is moving upward, a negative value means it is moving downward, and zero means it is momentarily stopped.

Visible text: The sign of tells the direction of motion at that moment. A positive value means the object is moving upward, a negative value means it is moving downward, and zero means it is momentarily stopped.

Notice that height $$y$$ does not automatically tell the motion direction. An object can be at the same height while moving upward and while moving downward, but the velocity signs are different.

Visible text: Notice that height does not automatically tell the motion direction. An object can be at the same height while moving upward and while moving downward, but the velocity signs are different.

| Value of $$v_y$$ | Motion Direction | Example Situation |
| --- | --- | --- |
| Positive | Upward | A ball just thrown upward |
| Zero | Stopped for an instant | A ball at the highest point |
| Negative | Downward | A ball falling back down |

Visible text: | Value of | Motion Direction | Example Situation |
| --- | --- | --- |
| Positive | Upward | A ball just thrown upward |
| Zero | Stopped for an instant | A ball at the highest point |
| Negative | Downward | A ball falling back down |

## A Ball Dropped From a Height

Suppose a ball is dropped from a height of $$20\text{ m}$$ with no initial velocity. With $$g=10\text{ m/s}^2$$, the time to reach the ground is:

Visible text: Suppose a ball is dropped from a height of with no initial velocity. With , the time to reach the ground is:

```math
\begin{aligned}
0&=20-\frac{1}{2}(10)t^2\\
5t^2&=20\\
t&=2\text{ s}
\end{aligned}
```

The velocity just before reaching the ground is:

```math
v_y=0-10(2)=-20\text{ m/s}
```

The negative sign shows the downward direction. So the speed is $$20\text{ m/s}$$, but the motion direction is downward.

Visible text: The negative sign shows the downward direction. So the speed is , but the motion direction is downward.

## From Vertical Motion to Projectile Motion

Vertical motion also appears in projectile motion. Later, when an object moves at an angle, the horizontal part is read as sideways motion, while the vertical part still follows the gravity equations from this lesson.

That is why free-fall motion is useful beyond this page. The vertical part of a thrown ball, a projectile, or water leaving a hose is still shaped by gravitational acceleration.