\(x_0\)
\(m\)
\(y_0\)
\(m\)
\(v_{0,x},v_{0,y}\)
\(m.s\)
\(g\)
\(m.s^{-2}\)



Projectile Motion

To determine to position of the projectile we should compute the position vector \(\vec{r}(t)=x(t)\vec{i}+y(t)\vec{i}\).

\(x(t)\):

We know from Newton second law that \(\sum \vec{F} = m\times \vec{a}_x = m\times a_x(t)\vec{i}\)

However, the projectile as a constant speed along \(\vec{i}\). Hence, \(a_x(t) = 0 \).

Thus:

\[ x(t) = \int_{t_0}^t v_{0,x}dt = v_{0,x}t + C = v_{0,x}t + x_0\]
\(y(t)\):

We know from Newton second law that \(\sum \vec{F} = m\times \vec{a}_y = m\times a_y(t)\vec{i}\)

The projectile is under the influence of the gravity that is oriented downward. Hence, \(a_y(t) = -g \).

Thus:

\[ v_y(t) = \int_{t_0}^t a_{y}(t)dt = -gt+C = -gt + v_{0,y}\] \[ y(t) = \int_{t_0}^t v_y(t)dt = -\frac{1}{2}gt^2 + v_{0,y}t+C=-\frac{1}{2}gt^2 + v_{0,y}t+y_0\]
\(\vec{r}(t)\):
Finally knowing \(x(t)\) and \(y(t)\) we have \( \vec{r}(t) = \left(\begin{smallmatrix}x(t)\\y(t)\end{smallmatrix}\right) = \left(\begin{smallmatrix}v_{0,x}t + x_0\\-\frac{1}{2}gt^2 + v_{0,y}t+y_0\end{smallmatrix}\right)\)