Non-Horizontally Launched Projectiles and their Trajectories

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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.

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:

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Components of initial velocity

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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} $.

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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.

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

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}