How would you derive the cartesian equation for projectile motion? (1 Viewer)

OreoMcFlurry · Aug 9, 2021

Hey guys,

How would you derive the cartesian equation for projectile motion from the horizontal and vertical accelerations of the object?

In other words, How would you derive the cartesian equation from this step:

x double dot = 0
&
y double dot = -g

(assuming that upwards is positive)

Life'sHard · Aug 9, 2021

x dot would just be vcostheta + c

y dot would be -gt + vsintheta + c

To find the constant the question would specify what is occuring at that velocity and time. Same with x and y displacement.

x = vtcostheta + c

y = (gt^2)/2 + vtsintheta + c

Valvesound · Aug 9, 2021

OreoMcFlurry said:
Hey guys,

How would you derive the cartesian equation for projectile motion from the horizontal and vertical accelerations of the object?

In other words, How would you derive the cartesian equation from this step:

x double dot = 0
&
y double dot = -g

(assuming that upwards is positive)

Do what Life'sHard did and then make t the subject in the x component and sub that into the y component

OreoMcFlurry · Aug 10, 2021

Life'sHard said:
x dot would just be vcostheta + c

y dot would be -gt + vsintheta + c

To find the constant the question would specify what is occuring at that velocity and time. Same with x and y displacement.

x = vtcostheta + c

y = (gt^2)/2 + vtsintheta + c

So, when integrating from acceleration to velocity (ie. x double dot and y double dot to x dot and y dot respectively), we substitute c with the initial horizontal and vertical velocities. Do we need to write "Initial conditions" to explain why c is the initial velocities or is it not required?

CM_Tutor · Aug 10, 2021

Explained slightly differently, the path equation links $x$ and $y$ . The above answers explain how to get from $\ddot{x}$ and $\ddot{y}$ to $x$ and $y$ as functions of $t$ , time. You then can find the path by eliminating $t$ because you have every point on the path as $\left(x(t),\ y(t)\right)$ and so $t$ is just a parameter.

For MX1 projectiles, you know the path is a (part of a) parabola and you can find a position at any particular moment by substituting $t$ , but to find properties of the path itself, you may need it in Cartesian rather than Parametric forms.

Conceptually, this is the same as taking $\left(r\cos\theta,\ r\sin\theta\right)$ as a point on a circle centred at $O$ and need to interchange with the Cartesian form $x^2 + y^2 = r^2$ , which you find by eliminating the parameter $\theta$ .

CM_Tutor · Aug 10, 2021

OreoMcFlurry said:
So, when integrating from acceleration to velocity (ie. x double dot and y double dot to x dot and y dot respectively), we substitute c with the initial horizontal and vertical velocities. Do we need to write "Initial conditions" to explain why c is the initial velocities or is it not required?

Suppose the particle is projected from the origin at time $t = 0$ at $V\ \text{m s}^{-1}$ at an angle $\alpha$ to the horizontal:

$\begin{align*} \ddot{x} &= 0 \\ \int \ddot{x}\ dt = \dot{x} &= C_1,\ \text{for some constant $C_1$} \\ \text{At $t = 0$, $\dot{x} = V\cos\alpha$:} \quad V\cos\alpha &= C_1 \\ \text{So,} \quad \dot{x} &= V\cos\alpha \\ \\ \int \dot{x}\ dt &= \int V\cos\alpha\ dt \\ x &= Vt\cos\alpha + C_2,\ \text{for some constant $C_2$} \\ \text{At $t = 0$, $x = 0$:} \quad 0 &= 0 \times V\cos\alpha + C_2 \\ 0 &= C_2 \\ \text{So,} \quad x &= Vt\cos\alpha \end{align*}$

$\begin{align*} \ddot{y} &= -g \\ \int \ddot{y}\ dt &= \int -g\ dt \\ \dot{y} &= -gt + C_3,\ \text{for some constant $C_3$} \\ \text{At $t = 0$, $\dot{y} = V\sin\alpha$:} \quad V\sin\alpha &= -g \times 0 + C_3 \\ V\sin\alpha &= C_3 \\ \text{So,} \quad \dot{y} &= -gt + V\sin\alpha \\ \\ \int \dot{y}\ dt &= \int -gt + V\sin\alpha\ dt \\ y &= -g\cfrac{t^2}{2} + Vt\sin\alpha + C_4\ \text{for some constant $C_4$} \\ \text{At $t = 0$, $y = 0$:} \quad 0 &= -\cfrac{g}{2} \times 0^2 + 0 \times V\sin\alpha + C_4 \\ 0 &= C_4 \\ \text{So,} \quad y &= -\cfrac{gt^2}{2} + Vt\sin\alpha \end{align*}$

OreoMcFlurry · Aug 10, 2021

CM_Tutor said:
Suppose the particle is projected from the origin at time $t = 0$ at $V\ \text{m s}^{-1}$ at an angle $\alpha$ to the horizontal:

$\begin{align*} \ddot{x} &= 0 \\ \int \ddot{x}\ dt = \dot{x} &= C_1,\ \text{for some constant $C_1$} \\ \text{At $t = 0$, $\dot{x} = V\cos\alpha$:} \quad V\cos\alpha &= C_1 \\ \text{So,} \quad \dot{x} &= V\cos\alpha \\ \\ \int \dot{x}\ dt &= \int V\cos\alpha\ dt \\ x &= Vt\cos\alpha + C_2,\ \text{for some constant $C_2$} \\ \text{At $t = 0$, $x = 0$:} \quad 0 &= 0 \times V\cos\alpha + C_2 \\ 0 &= C_2 \\ \text{So,} \quad x &= Vt\cos\alpha \end{align*}$

$\begin{align*} \ddot{y} &= -g \\ \int \ddot{y}\ dt &= \int -g\ dt \\ \dot{y} &= -gt + C_3,\ \text{for some constant $C_3$} \\ \text{At $t = 0$, $\dot{y} = V\sin\alpha$:} \quad V\sin\alpha &= -g \times 0 + C_3 \\ V\sin\alpha &= C_3 \\ \text{So,} \quad \dot{y} &= -gt + V\sin\alpha \\ \\ \int \dot{y}\ dt &= \int -gt + V\sin\alpha\ dt \\ y &= -g\cfrac{t^2}{2} + Vt\sin\alpha + C_4\ \text{for some constant $C_4$} \\ \text{At $t = 0$, $y = 0$:} \quad 0 &= -\cfrac{g}{2} \times 0^2 + 0 \times V\sin\alpha + C_4 \\ 0 &= C_4 \\ \text{So,} \quad y &= -\cfrac{gt^2}{2} + Vt\sin\alpha \end{align*}$

Thank you so much! I was confused on how to present such info when answering a projectile motion question and this has helped a lot

CM_Tutor · Aug 10, 2021

Be aware that MX1 methods now tend to treat projectile motion from a vector perspective, in which case the derivation is more like:

Suppose the particle is projected from the origin at time $t = 0$ at $V\ \text{m s}^{-1}$ at an angle $\alpha$ to the horizontal. Then the vector for the velocity of projection is $\overrightarrow{v_0} = V\left(\cos\alpha\overrightarrow{i} + \sin\alpha\overrightarrow{j}\right)$ . The vectors $\overrightarrow{a}$ , $\overrightarrow{v}$ , and $\overrightarrow{r}$ are the vectors for the acceleration, velocity, and position of the projectile at time $t$ .

$\begin{align*} \overrightarrow{a} &= \ddot{x}\overrightarrow{i} + \ddot{y}\overrightarrow{j} \\ &= -g\overrightarrow{j} \qquad \text{as $\ddot{x} = 0$ and $\ddot{y} = -g$} \\ \int \overrightarrow{a}\ dt &= \int -g\overrightarrow{j}\ dt \\ \overrightarrow{v} &= -gt\overrightarrow{j} + \overrightarrow{C_1},\ \text{for some constant vector $\overrightarrow{C_1}$} \\ \text{At $t = 0$, $\overrightarrow{v_0} = V\left(\cos\alpha\overrightarrow{i} + \sin\alpha\overrightarrow{j}\right)$:} \quad \overrightarrow{v_0} &= -g\overrightarrow{j} \times 0 + \overrightarrow{C_1} \\ \overrightarrow{v_0} &= \overrightarrow{C_1} \\ \text{So,} \quad \overrightarrow{v} &= -gt\overrightarrow{j} + \overrightarrow{v_0} \\ &= -gt\overrightarrow{j} + V\left(\cos\alpha\overrightarrow{i} + \sin\alpha\overrightarrow{j}\right) \\ &= V\cos\alpha\overrightarrow{i} + (V\sin\alpha - gt)\overrightarrow{j} \end{align*}$

$\begin{align*} \text{Now,} \qquad \int \overrightarrow{v}\ dt &= \int V\cos\alpha\overrightarrow{i} + (V\sin\alpha - gt)\overrightarrow{j} \ dt \\ \overrightarrow{r} &= Vt\cos\alpha\overrightarrow{i} + \left(Vt\sin\alpha - g\cfrac{t^2}{2}\right)\overrightarrow{j} + \overrightarrow{C_2},\ \text{for some constant vector $\overrightarrow{C_2}$} \\ \text{At $t = 0$, $\overrightarrow{r} = 0\overrightarrow{i} + 0\overrightarrow{j}$:} \quad 0\overrightarrow{i} + 0\overrightarrow{j} &= 0 \times V\cos\alpha\overrightarrow{i} + \left(V\sin\alpha \times 0 - g\cfrac{0^2}{2}\right)\overrightarrow{j} + \overrightarrow{C_2} \\ 0\overrightarrow{i} + 0\overrightarrow{j} &= \overrightarrow{C_2} \\ \text{So,} \qquad \overrightarrow{r} &= Vt\cos\alpha\overrightarrow{i} + \left(Vt\sin\alpha - \cfrac{gt^2}{2}\right)\overrightarrow{j} \end{align*}$

How would you derive the cartesian equation for projectile motion? (1 Viewer)

OreoMcFlurry

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Valvesound

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