Horizontally Launched Projectiles

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.

Experiment

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:

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.

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

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}

Velocity-time

The velocity along the x-axis \( v_0 \) is constant and the velocity along the y-axis increases uniformly because of the gravitational acceleration.

Uniform motion $$ v_x = v_0 = \rm konst. $$
Uniformly accelerated motion $$ v_y = - g \cdot t $$

The instantaneous velocity in the direction of flight is determined using the Pythagorean theorem of the velocity components.

$$ v(t) = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{(v_0)^2 + g^2 \cdot t^2 } $$

Sources