Detailed derivation of rectilinear velocity:

Assume we have rectilinear motion of a particle (of negligible size and shape, from A:pointMass); that is, motion in a straight line. The velocity is v and the acceleration is a. The motion in TM:acceleration is now one-dimensional with a constant acceleration, represented by a^c. The initial velocity (at t = 0, from A:timeStartZero) is represented by vⁱ. From TM:acceleration in 1D, and using the above symbols we have:

\[{a^{c}}=\frac{\,dv}{\,dt}\]

Rearranging and integrating, we have:

\[\int_{{v^{\text{i}}}}^{v}{1}\,dv=\int_{0}^{t}{{a^{c}}}\,dt\]

Performing the integration, we have the required equation:

\[v\text{(}t\text{)}={v^{\text{i}}}+{a^{c}} t\]

Detailed derivation of rectilinear position:

Assume we have rectilinear motion of a particle (of negligible size and shape, from A:pointMass); that is, motion in a straight line. The position is p and the velocity is v. The motion in TM:velocity is now one-dimensional. The initial position (at t = 0, from A:timeStartZero) is represented by pⁱ. From TM:velocity in 1D, and using the above symbols we have:

\[v=\frac{\,dp}{\,dt}\]

Rearranging and integrating, we have:

\[\int_{{p^{\text{i}}}}^{p}{1}\,dp=\int_{0}^{t}{v}\,dt\]

From GD:rectVel, we can replace v:

\[\int_{{p^{\text{i}}}}^{p}{1}\,dp=\int_{0}^{t}{{v^{\text{i}}}+{a^{c}} t}\,dt\]

Performing the integration, we have the required equation:

\[p\text{(}t\text{)}={p^{\text{i}}}+{v^{\text{i}}} t+\frac{{a^{c}} t^{2}}{2}\]

Detailed derivation of velocity vector:

For a two-dimensional Cartesian coordinate system (A:twoDMotion and A:cartSyst), we can represent the velocity vector as v(t) = \(\begin{bmatrix} {v_{\text{x}}}\\ {v_{\text{y}}} \end{bmatrix}\) and the acceleration vector as a(t) = \(\begin{bmatrix} {a_{\text{x}}}\\ {a_{\text{y}}} \end{bmatrix}\). The acceleration is assumed to be constant (A:constAccel) and the constant acceleration vector is represented as a^c = \(\begin{bmatrix} {{a_{\text{x}}}^{\text{c}}}\\ {{a_{\text{y}}}^{\text{c}}} \end{bmatrix}\). The initial velocity (at t = 0, from A:timeStartZero) is represented by vⁱ = \(\begin{bmatrix} {{v_{\text{x}}}^{\text{i}}}\\ {{v_{\text{y}}}^{\text{i}}} \end{bmatrix}\). Since we have a Cartesian coordinate system, GD:rectVel can be applied to each coordinate of the velocity vector to yield the required equation:

\[\symbf{v}\text{(}t\text{)}=\begin{bmatrix} {{v_{\text{x}}}^{\text{i}}}+{{a_{\text{x}}}^{\text{c}}} t\\ {{v_{\text{y}}}^{\text{i}}}+{{a_{\text{y}}}^{\text{c}}} t \end{bmatrix}\]

Detailed derivation of position vector:

For a two-dimensional Cartesian coordinate system (A:twoDMotion and A:cartSyst), we can represent the position vector as p(t) = \(\begin{bmatrix} {p_{\text{x}}}\\ {p_{\text{y}}} \end{bmatrix}\), the velocity vector as v(t) = \(\begin{bmatrix} {v_{\text{x}}}\\ {v_{\text{y}}} \end{bmatrix}\), and the acceleration vector as a(t) = \(\begin{bmatrix} {a_{\text{x}}}\\ {a_{\text{y}}} \end{bmatrix}\). The acceleration is assumed to be constant (A:constAccel) and the constant acceleration vector is represented as a^c = \(\begin{bmatrix} {{a_{\text{x}}}^{\text{c}}}\\ {{a_{\text{y}}}^{\text{c}}} \end{bmatrix}\). The initial velocity (at t = 0, from A:timeStartZero) is represented by vⁱ = \(\begin{bmatrix} {{v_{\text{x}}}^{\text{i}}}\\ {{v_{\text{y}}}^{\text{i}}} \end{bmatrix}\). Since we have a Cartesian coordinate system, GD:rectPos can be applied to each coordinate of the position vector to yield the required equation:

\[\symbf{p}\text{(}t\text{)}=\begin{bmatrix} {{p_{\text{x}}}^{\text{i}}}+{{v_{\text{x}}}^{\text{i}}} t+\frac{{{a_{\text{x}}}^{\text{c}}} t^{2}}{2}\\ {{p_{\text{y}}}^{\text{i}}}+{{v_{\text{y}}}^{\text{i}}} t+\frac{{{a_{\text{y}}}^{\text{c}}} t^{2}}{2} \end{bmatrix}\]

Detailed derivation of flight duration:

We know that p_yⁱ = 0 (A:launchOrigin) and a_y^c = −g (A:accelYGravity). Substituting these values into the y-direction of GD:posVec gives us:

\[{p_{\text{y}}}={{v_{\text{y}}}^{\text{i}}} t-\frac{g t^{2}}{2}\]

To find the time that the projectile lands, we want to find the t value (t_flight) where p_y = 0 (since the target is on the x-axis from A:targetXAxis). From the equation above we get:

\[{{v_{\text{y}}}^{\text{i}}} {t_{\text{flight}}}-\frac{g {t_{\text{flight}}}^{2}}{2}=0\]

Dividing by t_flight (with the constraint t_flight > 0) gives us:

\[{{v_{\text{y}}}^{\text{i}}}-\frac{g {t_{\text{flight}}}}{2}=0\]

Solving for t_flight gives us:

\[{t_{\text{flight}}}=\frac{2 {{v_{\text{y}}}^{\text{i}}}}{g}\]

From DD:speedIY (with vⁱ = v_launch) we can replace v_yⁱ:

\[{t_{\text{flight}}}=\frac{2 {v_{\text{launch}}} \sin\left(θ\right)}{g}\]

Detailed derivation of landing position:

We know that p_xⁱ = 0 (A:launchOrigin) and a_x^c = 0 (A:accelXZero). Substituting these values into the x-direction of GD:posVec gives us:

\[{p_{\text{x}}}={{v_{\text{x}}}^{\text{i}}} t\]

To find the landing position, we want to find the p_x value (p_land) at flight duration (from IM:calOfLandingTime):

\[{p_{\text{land}}}=\frac{{{v_{\text{x}}}^{\text{i}}}\cdot{}2 {v_{\text{launch}}} \sin\left(θ\right)}{g}\]

From DD:speedIX (with vⁱ = v_launch) we can replace v_xⁱ:

\[{p_{\text{land}}}=\frac{{v_{\text{launch}}} \cos\left(θ\right)\cdot{}2 {v_{\text{launch}}} \sin\left(θ\right)}{g}\]

Rearranging this gives us the required equation:

\[{p_{\text{land}}}=\frac{2 {v_{\text{launch}}}^{2} \sin\left(θ\right) \cos\left(θ\right)}{g}\]