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Horizontal Throw: How Bodies Fall in a Gravitational Field

Important Formula

Formula: Horizontal throw - parabolic path

$$y ~=~ -\frac{g}{ 2\,{v_0}^2 } \, x^2 ~+~ y_0$$ $$y ~=~ -\frac{g}{ 2\,{v_0}^2 } \, x^2 ~+~ y_0$$ $$x ~=~ v_0 \, \sqrt{ \frac{2 \, (y_0 ~-~ y) }{g} }$$ $$v_0 ~=~ x \, \sqrt{ \frac{g}{ 2 \, (y_0 ~-~ y) } }$$ $$y_0 ~=~ y ~+~ \frac{ g }{ 2 \, {v_0}^2 } \, x^2$$ $$g ~=~ \frac{ 2 \, (y_0 ~-~ y) }{ x^2 } \, {v_0}^2$$

Rearrange formula

What do the formula symbols mean?

Height

$$ y $$ Unit $$ \mathrm{m} $$

Current height $ y $ above the ground of a horizontally thrown body. $y$ is therefore the vertical distance of the body to the ground.

The given formula describes a parabolic trajectory $ y(x) $, which the body follows after being thrown / launched. Using this formula you can calculate the current height $y$ for each current horizontal position $x$ of the body.

This formula is a special case of the inclined throw. And the minus sign indicates that the body falls downwards (set as negative direction).

Horizontal distance

$$ x $$ Unit $$ \mathrm{m} $$

Distance of the body from the starting throw position to the current horizontal position $x$ of the body.

Initial velocity

$$ v_0 $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$

Horizontal velocity (in the body's $x$ direction) at the time of throwing (or releasing). This is the velocity of the body with which you threw the body horizontally.

It is assumed here that this velocity does not change during the flight, that is, it is constant at all times.

Initial height

$$ y_0 $$ Unit $$ \mathrm{m} $$

This is the height from which the body is released.

Gravitational acceleration

$$ g $$ Unit $$ \frac{\mathrm{m}}{\mathrm{s}^2} $$

A constant acceleration with the value $ g = 9.8 \, \frac{\text m}{\text{s}^2}$. The acceleration states that the thrown body increases its vertical velocity by $ 9.8 \, \frac{\text m}{\text{s}}$ every second. The body is in free fall after release.

Hover the image!

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 $.

Hover the image!

Trajectory curve of a body when you throw it horizontally.

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:

$$ \begin{align} s ~=~ \frac{1}{2} \, a \, t^2 ~+~ v_0 \, t ~+~ s_0 \end{align} $$

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:

$$ \begin{align} a_{\text y} ~=~ - g \end{align} $$

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:

$$ \begin{align} y ~=~ -\frac{1}{2} \, g \, t^2 ~+~ v_{\text y} \, t ~+~ y_0 \end{align} $$

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:

$$ \begin{align} y ~=~ -\frac{1}{2} \, g \, t^2 ~+~ y_0 \end{align} $$

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 $t$:

$$ \begin{align} x ~=~ \frac{1}{2} \, a_{\text x} \, t^2 ~+~ v_0 \, t ~+~ x_0 \end{align} $$

With the above considerations, the first and last summands in the distance-time law 5 cancel out and we get:

$$ \begin{align} x ~=~ v_0 \, t \end{align} $$

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 $:

$$ \begin{align} t ~=~ \frac{ x }{ v_0 } \end{align} $$

Substitute this equation into equation 4 for $ t $ to eliminate $ t $:

$$ \begin{align} y ~=~ -\frac{1}{2} \, g \, \left( \frac{ x }{ v_0 } \right)^2 ~+~ y_0 \end{align} $$

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:

$$ \begin{align} y(t) ~=~ -\frac{1}{2} \, g \, t^2 ~+~ y_0 \end{align} $$

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?

$$ \begin{align} y(t_{\text d}) ~=~ -\frac{1}{2} \, g \, {t_{\text d}}^2 ~+~ y_0 \end{align} $$

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:

$$ \begin{align} 0 ~=~ -\frac{1}{2} \, g \, {t_{\text d}}^2 ~+~ y_0 \end{align} $$

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

$$ \begin{align} \frac{1}{2} \, g \, {t_{\text d}}^2 &~=~ y_0 \\\\
{t_{\text d}}^2 &~=~ \frac{2 \, y_0}{g} \\\\ \end{align} $$

And the last rearranging step results:

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

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.

Example: Calculate flight duration of a projectile

A man shoots a pistol bullet horizontally at shoulder height (1.7 meters). When does the pistol bullet land on the ground?
For this we insert the gravitational acceleration $ g = 9.8 \, \frac{ \mathrm m }{ \mathrm{s}^2 } $ and the initial height $ y_0 = 1.7 \, \mathrm{m} $ into the throw duration formula:

$$ \begin{align} t_{\text d} &~=~ \sqrt{ \frac{ 2 \cdot 1.7 \, \mathrm{m} }{ 9.8 \, \frac{ \mathrm m }{ \mathrm{s}^2}} } \\\\
&~=~ 0.59 \, \mathrm{s} \end{align} $$

After 0.59 seconds, the fired pistol bullet lands on the ground, regardless of how heavy or how fast it is!

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 $:

$$ \begin{align} x(t) ~=~ v_0 \, t \end{align} $$

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} $):

$$ \begin{align} x(t_{\text d}) ~=~ v_0 \, t_{\text d} \end{align} $$

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

$$ \begin{align} w ~=~ v_0 \, t_{\text d} \end{align} $$

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:

$$ \begin{align} w ~=~ v_0 \, \sqrt{ \frac{2 \, y_0}{g} } \end{align} $$

Example: Calculate flight distance

How far will the pistol bullet land if it is launched horizontally from height $ 1.7 \, \mathrm{m} $ with velocity $ 800 \, \frac{ \mathrm m }{ \mathrm s} $?
Substitute the gravitational acceleration $ 9.8 \, \frac{ \mathrm m }{ \mathrm{s}^2 } $, the initial height $ 1.7 \, \mathrm{m} $, and the initial velocity $ 800 \, \frac{ \mathrm m }{ \mathrm s} $ into the formula for flight distance:

$$ \begin{align} w &~=~ 800 \, \frac{ \mathrm m }{ \mathrm s} \, \sqrt{ \frac{ 2 \cdot 9.8 \, \frac{ \mathrm m }{ \mathrm{s}^2 } }{ 1.7 \, \mathrm{m} } } \\\\
&~=~ 2716 \, \mathrm{m} \end{align} $$

So the pistol bullet lands about 2.7 kilometers away from where the bullet is fired.

fix

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