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h2. Part A

Suppose you are throwing a baseball.  You release the ball with a perfectly horizontal velocity of 5.0 m/s at a height of 1.5 m above the ground.  How far will the ball travel horizontally from the instant it leaves your hand until the instant it first contacts the ground?  

System:  The ball will be treated as a [point particle] subject to an influence from the earth (gravity).

Models:  The ball is in projectile motion, so we model the x-component of the ball's motion as [One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)] and the y-component as [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)].

Approach:  The first thing to do is to sketch the situation, which allows us to summarize the givens and unknowns and also to set up a coordinate system.  

!baseball2.png!

In the problem statement, we are told that _h_ = 1.5 m (as drawn in the picture) and we are asked for _d_.  By drawing coordinate axes into our picture we have denoted the positive _x_ and _y_ directions.  We have not yet chosen the origin, however (the axes can be placed wherever you wish on the picture to avoid clutter).  We will take that step now.  We choose our origin such that the position _x_ = 0 m is the location at which the ball leaves the hand.  The location _y_ = 0 m is the level of the ground.  With these choices 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 a sign of a math error!  Such warnings are extremely valuable when checking work.{note}

Part B

Suppose you are throwing a baseball.  You release the ball with a perfectly horizontal velocity at a height of 1.5 m above the ground.  The ball travels 5.0 m horizontally from the instant it leaves your hand until the instant it first contacts the ground.  How fast was the ball moving when you released it?

System & Models:  As in Part A.

Approach:  This question works basically the same as Part A, except that the givens are slightly different.  In this part, we have (assuming the same coordinate system as was used in Part A):
{panel:title=givens}{latex}\begin{large}\[ t_{\rm i} = \mbox{0 s} \] \[ x_{\rm i} = \mbox{0 m} \]\[ x = d = \mbox{5.0 m}\] \[ y_{\rm i} = \mbox{1.5 m} \] \[ y = \mbox{0 m}\]\[v_{y,{\rm i}} = \mbox{0 m/s} \] \[ a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}{latex}{panel}
{note}Note that although we are not given the _x_ component of the initial velocity, the phrase "perfectly horizontal velocity" still tells us that the _y_ component is zero initially.{note}

Once again, we have an unknown that requires the _x_ direction's Law of Change, but we do not have enough information.  We proceed in exactly the same fashion as in Part A to find the same answer for the time:

{latex}\begin{large} \[ t = \sqrt{\frac{-2y_{\rm i}}{a_{y}}} \] \end{large}{latex}

We then rearrange the _x_ direction equation and substitute for the time:

{latex}\begin{large} \[ v_{x} = \frac{x}{t} = x \sqrt{\frac{a_{y}}{-2y_{\rm i}}} = (\mbox{5.0 m})\sqrt{\frac{-\mbox{9.8 m/s}^{2}}{-2(\mbox{1.5 m})}} = \mbox{9.0 m/s} \] \end{large}{latex}

Part C