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