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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}
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With the choice of coordinates made, we can summarize the givens (along with our traditional choice that _t_~i~ = 0 s):
{panel:title=givens}{latex}\begin{large}\[ t_{\rm i} = \mbox{0 s} \] \[ x_{\rm i} = \mbox{0 m} \]\[ x = d \] \[ y_{\rm i} = \mbox{1.5 m} \] \[ y = \mbox{0 m}\]\[v_{x} = \mbox{5.0 m/s} \]\[v_{y,{\rm i}} = \mbox{0 m/s} \] \[ a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}{latex}{panel}
{note}It is important to note that the phrase *perfectly horizontal velocity of 5.0 m/s* implies that the full velocity (5.0 m/s) is directed along the x-axis, with zero y-component for the initial velocity. This phrasing is extremely common in physics. You will also encounter the perpendicular case of a "perfectly vertical velocity". It is also worth remarking that although 5.0 m/s is the velocity at the instant of release (clearly the ball's initial velocity for the freefall trajectory of interest) we have written _v_~_x_~ = 5.0 m/s rather than _v_~_x_,i~ = 5.0 m/s. This is not a typo, because the _x_ direction is subject to the 1-D Motion with Constant Velocity model (recall _a_~_x_~ = 0). Because the _x_ velocity is constant, it does not require labels for initial or final states.{note}
We are asked for _d_, which appears in the _x_ direction givens. For this reason, we should first consider the Laws of Change available for the _x_ direction. Because of the simplicity of the 1-D Motion with Constant Velocity model, there is only one available Law of Change:
{latex}\begin{large} \[ x = x_{\rm i} + v_{x} (t-t_{\rm i}) \] \end{large}{latex}
We cannot solve this equation, however, because we do not know _x_ or _t_. We therefore proceed to use an extremely useful technique. We will use the _y_ direction to solve for the time _t_. Once we have this, we can use it in the _x_ direction equation to obtain the information requested by the problem.
{note}In projectile motion, if the direction which explicitly contains the desired unknown quantity does not yield solvable equations using the problem givens, it is *extremely* likely that you should solve the other direction for time.{note}
In the _y_ direction we have four Laws of Change to consider. The most direct way to proceed is to use:
{latex}\begin{large} \[ y = y_{\rm i} + v_{y,{\rm i}}(t-t_{\rm i}) + \frac{1}{2} a_{y}(t-t_{\rm i})^{2}\]\end{large}{latex}
which at first looks messy, but after substituting zeros becomes:
{latex}\begin{large} \[ 0 = y_{\rm i} + \frac{1}{2} a_{y}t^{2} \] \end{large}{latex}
which is solved to give:
{latex}\begin{large} \[ t = \pm \sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}{latex}
and we must choose the plus sign since we have already set up the problem with the ball released at _t_ = 0 s.
This time can be substituted directly into the _x_ direction Law of Change to give:
{latex}\begin{large} \[ x = d = v_{x}\sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}{latex}
To be clear, we show the substitution:
{latex}\begin{large} \[ d = (\mbox{5.0 m/s})\sqrt{\frac{-2(\mbox{1.5 m})}{-\mbox{9.8 m/s}^{2}}} = \mbox{2.8 m} \] \end{large}{latex}
{note}Note that the negative sign under the square root was canceled by the negative _y_ acceleration. When you see a negative sign appear under a square root, you should always check that it is canceled by the algebraic signs of the given quantities. If it does not cancel, it is |