﻿ Horizontally Launched Projectiles - College Physics

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.

ResetStart
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

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