﻿ Vertically Launched Projectiles - College Physics

# Vertically Launched Projectiles

## Introduction

The vertical launch of projectiles represents an superimposition of straight uniform motion upward and the free fall.

## Experiment

A projectile is launched with the initial velocity $$v_0$$ upwards. It initially rises quickly, but slows down until it has reached the highest point. Then it slowly begins to fall and continues until it hits the ground.

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Legende
Geschwindigkeit
Beschleunigung

## Results

The horizontal launch is a uniformly accelerated motion with the constant acceleration due to gravity $$g$$ and the initial velocity $$v_0$$. Therefore the position, velocity and acceleration of the projectile during the flight can be described by the following formulas:

$$y = v_0 \cdot t - \dfrac{g}{2} \cdot t^2$$ Distance-time $$v = v_0 - g \cdot t$$ Velocity-time $$a = g = \rm 9,81 \,\, \frac{m}{s^2} = \rm{konst.}$$ Acceleration-time

## 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 &= v_0 - g \cdot t \\ \\ 0 &= v_0 - g \cdot t_\rm{H} \\ \\ v_0 &= g \cdot t_\rm{H} \\ \\ t_\rm{H} &= \dfrac{v_0}{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} &= v_0 \cdot t_\rm{H} - \dfrac{g}{2} \cdot (t_\rm{H})^2 \\ \\ y_\rm{max} &= v_0 \cdot \dfrac{v_0}{g} - \dfrac{g}{2} \cdot \left(\dfrac{v_0}{g}\right)^2 \\ \\ y_\rm{max} &= \dfrac{(v_0)^2}{g} - \dfrac{\cancel g}{2} \cdot \dfrac{(v_0)^2}{g^{\cancel 2}} \\ \\ y_\rm{max} &= \dfrac{(v_0)^2}{g} - \dfrac{1}{2} \cdot \dfrac{(v_0)^2}{g} \\ \\ y_\rm{max} &= \dfrac{(v_0)^2}{2 \,\, g} \\ \\ \end{aligned}

## Distance-time equation

Using the position-time equation we can determine the height of the projectile dependant of the time. But for determining the traveled distance we need the distance-time equation:

$$s = v_0 \cdot t - \dfrac{g}{2} \cdot t^2$$ Distance-time equation for $$t \leq t_\rm{H}$$ $$s = s_\rm{max} + \dfrac{g}{2} \cdot \left(t-t_\rm{H}\right)^2$$ Distance-time equation for $$t \gt t_\rm{H}$$

### Sources

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