# Horizontal throw: How bodies fall parabolically

**Level 2**requires school mathematics. Suitable for pupils.

## Table of contents

In the following we want to look at a throw where the body is dropped *horizontally* with a *constant* **initial velocity** \( v_0 \) in the \(x\) direction from an **initial height** \( y_0 \).

A horizontal (horizontal) throw represents a *two-dimensional* movement. The thrown body not only moves horizontally, but also falls in vertical direction to the ground. Therefore, we need a two-dimensional coordinate system for the study of the horizontal throw.

The movement along the \( x \) axis describes a horizontal movement.

The movement along the \( y \) axis describes a vertical movement.

We can consider the vertical and horizontal movement independently from each other.

The motion of a body that experiences **acceleration** \( a \), has **initial velocity** \( v_0 \), and starts at **initial position** \( s_0 \) is generally given by the following Distance-Time law:

**Distance-Time law**

In this way we can calculate the current position \( s \) of the body at any time \( t \). Of course, the distance-time law applies to both the vertical and horizontal motion of the body. Let's apply it to the vertical motion first.

## Vertical movement after a horizontal throw

Let us consider only the vertical motion of the body. The body is attracted by the earth and thus experiences a **gravitational acceleration** \( g \) downwards. We define the direction 'down' as negative direction and the motion 'up' as positive \(y\) direction (see illustration 1).

The **vertical acceleration** \( a_{\text y} \) in the \(y\) direction is thus:

**Vertical acceleration is negative gravitational acceleration**

We have thrown the body exactly in the horizontal direction, which in turn means that the **vertical initial velocity** \( v_{\text y0} \) is zero: \( v_{\text y0} = 0\).

Let us now adjust the distance-time law 1

for the vertical motion:

The \( s \) corresponds to the current height \( y \) above the ground at time \( t \).

The \( a \) corresponds to the vertical acceleration \( a_{\text y} \) or, according to Eq.

2

, to the negative gravitational acceleration \( - g \).The \( v_0 \) corresponds to the initial vertical velocity \( v_{\text y0} \).

The \( s_0 \) corresponds to the initial height \( y_0 \) above the ground.

Thus, our distance-time law adapted for vertical motion is:

**Vertical position of the body**

In our case, we assumed that the body had no initial vertical velocity, so we set \( v_{\text y} = 0 \) in the equation. This eliminates the middle summand:

**Current vertical position (height) of the body**

## Horizontal movement

Next, we look at horizontal motion only. We again use the distance-time law 1

and adapt it for the horizontal motion:

The \( s \) corresponds to the current horizontal position \( x \) at time \( t \).

The \( a \) corresponds to the horizontal acceleration \( a_{\text x} \), which is zero in our case: \( a_{\text x} = 0 \).

The \( v_0 \) corresponds to the initial horizontal velocity \( v_{\text x0} \), which we refer to simply as \( v_0 \).

The \( s_0 \) corresponds to the start position \( x_0 \). We have placed the coordinate system so that \( x_0 = 0 \).

This gives us the adjusted distance-time law, which allows us to specify the horizontal position \(x\) of the body at each time \(t\):

**General formula for the horizontal position during a horizontal throw**

With the above considerations, the first and last summands in the distance-time law 5

cancel out and we get:

**Horizontal position of the body**

Now we can combine both equations 4

and 6

and thus eliminate the unknown time \( t \). For this purpose, rearrange the equation 6

of the horizontal motion with respect to time \( t \):

**Time is distance divided by speed**

Substitute this equation into equation 4

for \( t \) to eliminate \( t \):

**Trajectory curve for horizontal throw as a function of horizontal position**

We can exploit this equation whenever no time \( t \), such as the duration of the throw, is given in a task. As you can see from the formula, the current height \( y \) depends quadratically on the horizontal position \( x \). This in turn means that the projectile motion is **parabolic!**

Next, we want to figure out some important quantities, such as the throw duration and throw distance, to describe the throw more accurately.

## How long does a throw last?

Since we can consider vertical and horizontal motion independently, we exploit the vertical motion to find out the duration of the throw. Considered separately, the vertical motion represents a free fall. That is, to determine the duration of the throw, we need to find out how long the body falls to the ground.

Let's denote the **throw duration** (also called **throw time** or more generally **flight duration**) by \( t_{\text d} \).

Let us use the adjusted distance-time law 4

for the vertical motion of the body:

**Height as a function of time**

We have still notated here the dependence of \( t \) to make clear that it is a function \( y \) depending on the time \( t \). This equation tells us at what height \( y(t) \) the body is at time \( t \). That means we have to ask ourselves first:

*What is the vertical position \( y(t_{\text d}) \) of the body after the throw time \( t_{\text d} \) has passed?*

**Distance-time law for the throw duration**

This is not difficult to answer, because the throw time \( t_{\text d} \) represents the time after which the body has landed on the ground. And the ground has the vertical position \( y = 0 \). Thus, because of \( y(t_{\text d}) = 0 \), we can set the left side of 10

equal to zero:

**Distance-time law for the throw duration set equal to zero**

Rearrange it with respect to the throw time \( t_{\text d} \):

**Rearranged distance-time law for the throw duration**

{t_{\text d}}^2 &~=~ \frac{2 \, y_0}{g} \\\\ \end{align} $$

And the last rearranging step results:

**Formula for the flight duration (throw time)**

Very nice! To find out the throw time, we only need to know the initial height \( y_0 \) from which the body is thrown/shot.

## How far does the body fly?

To find out how far the thrown body lands from the initial horizontal position \( x = 0 \), we need to determine the **throw distance** (**flight distance**) \( w \). In this case, only the *horizontal* motion of the body is relevant. Its current height does not matter.

We know that the body flies the time \( t_{\text d} \) before it lands on the ground. Within this time the body moves in horizontal direction, which represents the distance from the starting position. The body moves with a constant velocity \( v_0 \) in the horizontal direction. We know according to Eq. 6

that after the time \( t \) this body covers the distance \( x \):

**Horizontal distance as a function of time**

If we substitute the throw time \( t_{\text d} \) for \( t \), we get the farthest possible position \( x(t_{\text d}) \) of the body (because the body is already on the ground after the time \( t_{\text d} \)):

**Distance-time law for throw distance**

The farthest horizontal position \( x(t_{\text d}) \) reached by the body corresponds exactly to the throw distance \( w \):

**Formula: Throw distance using throw duration**

If the duration of the throw is not given directly, but the initial height \( y_0 \), then we can substitute the throw duration formula 13

into 17

for it. We get:

**Formula for the flight distance (throw distance)**