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.
ResetStart
Legende
Geschwindigkeit
Beschleunigung
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-timeDistance-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
- Wikipedia: Article about "Trajectory of a projectile"
- Wikipedia: Article about "Range of a projectile"
- Deutsche Version: Artikel über "Senkrechter Wurf"