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      <div class="input-group-text">\(x_0\)</div>
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<br /><br /><br />

<h3>Projectile Motion</h3>
<p>To determine to position of the projectile we should compute the position vector \(\vec{r}(t)=x(t)\vec{i}+y(t)\vec{i}\).</p>
<h5>\(x(t)\):</h5>
<p>We know from Newton second law that \(\sum \vec{F} = m\times \vec{a}_x = m\times a_x(t)\vec{i}\)</p>
<p>However, the projectile as a constant speed along \(\vec{i}\). Hence, \(a_x(t) = 0 \).</p>
<p>Thus:</p>
\[ x(t) = \int_{t_0}^t v_{0,x}dt = v_{0,x}t + C = v_{0,x}t + x_0\]
<h5>\(y(t)\):</h5>
<p>We know from Newton second law that \(\sum \vec{F} = m\times \vec{a}_y = m\times a_y(t)\vec{i}\)</p>
<p>The projectile is under the influence of the gravity that is oriented <em>downward</em>. Hence, \(a_y(t) = -g \).</p>
<p>Thus:</p>
\[ v_y(t) = \int_{t_0}^t a_{y}(t)dt = -gt+C = -gt + v_{0,y}\]
\[ y(t) = \int_{t_0}^t v_y(t)dt = -\frac{1}{2}gt^2 + v_{0,y}t+C=-\frac{1}{2}gt^2 + v_{0,y}t+y_0\]
<h5>\(\vec{r}(t)\):</h5>
Finally knowing \(x(t)\) and \(y(t)\) we have \( \vec{r}(t) = \left(\begin{smallmatrix}x(t)\\y(t)\end{smallmatrix}\right) = \left(\begin{smallmatrix}v_{0,x}t + x_0\\-\frac{1}{2}gt^2 + v_{0,y}t+y_0\end{smallmatrix}\right)\)