Horizontally Launched Projectiles

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Non-Horizontally Launched Projectiles and their Trajectories

Introduction

The horizontal launch of a projectile is when it is thrown parallel to the horizon, so it moves with a horizontal takeoff speed only under the influence of its own weight.

A projectile is launched horizontally on a hill ($ h_0 = \rm 80 \,\, m $) with the initial velocity $ v_0 = \rm 40 \,\, \frac{m}{s} $. It moves in launch direction while falling faster and faster towards the ground.

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Legende
Geschwindigkeit
Beschleunigung

Results

Launching a projectile horizontally results in a combination of a uniform motion along the x-axis and a uniformly accelerated motion along the y-axis. Thus the movement (trajectory) can be presented in a $y(x) $ diagram:

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Determining the trajectory

To derive the trajectory the following laws are needed:

Uniform motion $$ x = v_0 \cdot t $$

Uniformly accelerated motion $$ y = h_0 - \dfrac{g}{2} \cdot t^2 $$

Now, the equation for the x-axis is solved for $t $ and put into the equation for the y-axis:

$$ x = v_0 \cdot t \qquad \Rightarrow \qquad t = \dfrac{x}{v_0} $$ \begin{aligned} y &= h_0 - \dfrac{g}{2} \cdot t^2 \\ \\ y &= h_0 - \dfrac{g}{2} \cdot \left( \dfrac{x}{v_0} \right)^2 \\ \\ y &= h_0 - \dfrac{g}{2} \cdot \dfrac{x^2}{(v_0)^2} \\ \\ y(x) &= h_0 - \dfrac{g}{2 \,\, (v_0)^2} \cdot x^2 \\ \\ \end{aligned}

Characteristics

Using the following graphs we can determine some characteristics of the horizontal launch. In the graph $s(x) $ on the left, the trajectory (see above) and the maximum range are shown. In graph on the right $s(t) $ the falling time is shown.

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$$ y(x) = h_0 - \dfrac{g}{2 \,\, (v_0)^2} \cdot x^2 $$

Maximum range

The maximum range is reached when the body hits the ground, that is when $y(x) $ is equal to zero:

\begin{aligned} y(x) &= h_0 - \dfrac{g}{2 \,\, (v_0)^2} \cdot x^2 \\ \\ 0 &= h_0 - \dfrac{g}{2 \,\, (v_0)^2} \cdot (x_\rm{max})^2 \\ \\ h_0 &= \dfrac{g}{2 \,\, (v_0)^2} \cdot (x_\rm{max})^2 \\ \\ \dfrac{2 \,\, (v_0)^2 \,\, h_0}{g} &= (x_\rm{max})^2 \\ \\ x_\rm{max} &= v_0 \cdot \sqrt{ \dfrac{2 \,\, h_0}{g} } = v_0 \cdot t_\rm{F} \\ \\ \end{aligned}

$$ y(t) = h_0 - \dfrac{g}{2} \cdot t^2 $$

Falling time

The body falls until it hits the ground, that is $y(t) $ is equal to zero:

\begin{aligned} y(t) &= h_0 - \dfrac{g}{2} \cdot t^2 \\ \\ 0 &= h_0 - \dfrac{g}{2} \cdot (t_\rm{F})^2 \\ \\ h_0 &= \dfrac{g}{2} \cdot (t_\rm{F})^2 \\ \\ \dfrac{2 \,\, h_0}{g} &= (t_\rm{F})^2 \\ \\ t_\rm{F} &= \sqrt{ \dfrac{2 \,\, h_0}{g} } \\ \\ \end{aligned}