﻿ Non-Horizontally Launched Projectiles and their Trajectories - College Physics

# Non-Horizontally Launched Projectiles and their Trajectories

## Introduction

A projectile is launched with a given angle to the horizontal. The resulting motion is a combination of uniform motion along the x-axis and free fall.

## Experiment

A projectile is launched from a hill ($$h_0 = \rm 30 \,\, m$$) with the initial velocity $$v_0 = \rm 40 \,\, \frac{m}{s}$$ at an angle $$\alpha = 20^\circ$$. It initially travels upwards until it reaches its maximum height and then falls faster and faster towards the ground.

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Legende
Geschwindigkeit
Beschleunigung

## Results

Launching a projectile non-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:

## Components of initial velocity

The initial velocity $$v_0$$ is divided, depending on launch angle $$\ alpha$$ into the components $$v_x$$ and $$v_y$$:

$$v_0 = \sqrt{ (v_x)^2 + (v_y)^2 }$$ $$v_{0,x} = v_0 \cdot \cos \alpha$$ $$v_{0,y} = v_0 \cdot \sin \alpha$$

## Determining the trajectory

To derive the trajectory the following laws are needed:

Uniform motion $$x = v_0 \cdot \cos \alpha \cdot t$$
Uniformly accelerated motion $$y = h_0 - \dfrac{g}{2} \cdot t^2 + v_0 \cdot \sin \alpha \cdot t$$

Now, the equation for the x-axis is solved for $$t$$ and used in the equation for the y-axis:

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

## Distance-time curve

The distance-time curve is a parabola which is shown in the following graph. In this graph you can also see the rise time $$t_\rm{H}$$ and the maximum height $$y_\rm{max}$$.

### Rise time

The body moves upwards until its speed along the y-axis is equal to zero, then it falls. Therefore solving the velocity-time equation for the velocity zero, the result is the rise time $$t_\rm{H}$$:

\begin{aligned} v_y &= v_0 \cdot \sin \alpha - g \cdot t \\ \\ 0 &= v_0 \cdot \sin \alpha - g \cdot t_\rm{H} \\ \\ v_0 \cdot \sin \alpha &= g \cdot t_\rm{H} \\ \\ t_\rm{H} &= \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ \end{aligned}

### Maximum height

After the rise time $$t_\rm{H}$$ the body has reached the maximum height. By putting the above formula for the rise time into the distance-time equation we obtain the maximum height $$y_\rm{max}$$:

\begin{aligned} y_\rm{max} &= y(t_\rm{H}) \\ \\ y_\rm{max} &= h_0 - \dfrac{g}{2} \cdot (t_\rm{H})^2 + v_0 \cdot \sin \alpha \cdot t_\rm{H} \\ \\ y_\rm{max} &= h_0 - \dfrac{g}{2} \cdot \left(\dfrac{v_0 \cdot \sin \alpha}{g}\right)^2 + v_0 \cdot \sin \alpha \cdot \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ y_\rm{max} &= h_0 - \dfrac{\cancel g}{2} \cdot \dfrac{(v_0 \cdot \sin \alpha)^2}{g^{\cancel 2}} + \dfrac{(v_0 \cdot \sin \alpha)^2}{g} \\ \\ y_\rm{max} &= h_0 - \dfrac{1}{2} \cdot \dfrac{(v_0 \cdot \sin \alpha)^2}{g} + \dfrac{(v_0 \cdot \sin \alpha)^2}{g} \\ \\ y_\rm{max} &= h_0 + \dfrac{(v_0)^2 \cdot (\sin \alpha)^2}{2 \,\, g} \\ \\ \end{aligned}

The rise time and thus the height become maximal when the launch angle $$\alpha$$ is $$90^\circ$$, as $$\sin 90^\circ = 1$$.

## Maximum range for $$h_0 = 0$$

The maximum range for $$h_0 = 0$$ can be derived easily. The following graph shows the trajectory of a projectile with the initial velocity $$v_0 = \rm 40 \,\, \frac{m}{s}$$ and the launch angle $$\alpha = 40^\circ$$. The maximum range is highlighted.

$$y(x) = \dfrac{g}{2 \,\, (v_0)^2} \cdot x^2$$ $$x(t) = v_0 \cdot \cos \alpha \cdot t \qquad \qquad \qquad y(t) = \dfrac{g}{2} \cdot t^2 + v_0 \cdot \sin \alpha \cdot t$$

The maximum range is reached when the time $$t_1 = t_\rm{H} + t_\rm{F}$$ (rise time + fall time) has elapsed. As the body falls the same time as it rises it holds $$t_\rm{F} = t_\rm{H}$$. The formula for the rise time was derived above.

\begin{aligned} x_\rm{max} &= x(2 \cdot t_\rm{H}) \\ \\ x_\rm{max} &= v_0 \cdot \cos \alpha \cdot 2 \cdot t_\rm{H} \\ \\ x_\rm{max} &= v_0 \cdot \cos \alpha \cdot 2 \cdot \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ x_\rm{max} &= (v_0)^2 \cdot 2 \cdot \dfrac{\cos \alpha \cdot \sin \alpha}{g} \qquad | \cos \alpha \cdot \sin \alpha = \dfrac{1}{2} \cdot \sin (2 \,\, \alpha)\\ \\ x_\rm{max} &= \dfrac{(v_0)^2 \sin (2 \,\, \alpha)}{g} \\ \\ \end{aligned}

## Velocity-time curve

The velocity along the x-axis is constant and the same as the inital velocity along x-axis $$v_{0, x}$$. The velocity along the y-axis is uniformly accelerated due to gravity.

Uniform motion $$v_x = v_{0,x} = v_0 \cdot \cos \alpha = \rm konst.$$
Uniformly accelerated motion $$v_y = v_{0,y} - g \cdot t = v_0 \cdot \sin \alpha - g \cdot t$$

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

\begin{aligned} v(t) &= \sqrt{ (v_x)^2 + (v_y)^2 } \\ \\ v(t) &= \sqrt{ (v_0 \cdot \cos \alpha)^2 + (v_0 \cdot \sin \alpha - g \cdot t)^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 \cdot (\cos \alpha)^2 + (v_0)^2 \cdot (\sin \alpha)^2 - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t + g^2 \cdot t^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 \cdot \underset{=1}{\underbrace{ ((\cos \alpha)^2 + (\sin \alpha)^2) }} - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t + g^2 \cdot t^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 + g^2 \cdot t^2 - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t } \\ \\ \end{aligned}

### Sources

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