\begin{large}\[ t_{\rm i} = \mbox{0 s} \] \[ x_{\rm i} = \mbox{0 m} \]\[ x = d \] \[ y_{\rm i} = \mbox{1.5 m} \] \[ y = \mbox{0 m}\]\[v_{x} |
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} = \mbox{5.0 m/s} \]\[v_{y,{\rm i}} = \mbox{0 m/s} \] \[ a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large} |
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{latex}| Note |
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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 vx = 5.0 m/s rather than vx,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 ax = 0). Because the x velocity is constant, it does not require labels for initial or final states. |
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 |
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\begin{large} \[ x = x_{\rm i} + v_{x} (t-t_{\rm i}) \] \end{large} |
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 |
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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. |
In the y direction we have four Laws of Change to consider. The most direct way to proceed is to use: | Latex |
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\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} |
which at first looks messy, but after substituting zeros becomes: | Latex |
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\begin{large} \[ 0 = y_{\rm i} + \frac{1}{2} a_{y}t^{2} \] \end{large} |
which is solved to give: | Latex |
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\begin{large} \[ t = \pm \sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large} |
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 |
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\begin{large} \[ x = d = v_{x}\sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large} |
To be clear, we show the substitution: | Latex |
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\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} |
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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 an indication of a math error! Such warnings are extremely valuable when checking work. |
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