Continuing with our projectile motion problems
this problems follows pretty much the same way all other projectile motion problems goes 
except it showcases some of the more advanced mathematical tools
that we may need to solve some of theparticular unknowns
 the setup here is fairly straightforward
while the math is a little tricky, but
once I show you the tools we have at our disposal it hopefully wouldn't be too bad
let's make sense of the situation we have with a little drawing
we have a tennis player serving
the tennis ball, so say he is back here, he has his racket up in the air
and he is [serving]
with a pretty massive velocity
initial velocity
at a certain angle below the horizontal as they say
and that it goes down
and the question is, first of all, does it clear the net, which is over here
and then secondly, does it lands inside or outside the serving box
so we want to know where it lands
we are given some numbers here that is good to keep track of
so at time = t_0, that's one point we're interested in
we know that it is 2.5 m up in the air
and then later, just as it goes over the net, we are interested at that point, which we know is
11.9 m between the nets and service line, and that the heights of the net is 0.91 m
then there is another time here we want to know about when it lands
we don't know any of these times, so we will use t_0, t_1 and t_2 as our placeholders for now
and once again we will write the kinematics equations for the x and the y separately 
and then solve for the appropriate numbers 
before we can do that though, it's important to set up a co-ordinate system
 and so let's call this point x = 0 and have x being horizontal, but let's set [the positive] up that way
because I have already drawn the tennis players serving from right to left
so that's my x, and then we will keep on having positive y upwards and we'll call y = 0 at the ground
so very critically, my y_0 not 0 this time, it is up a 2.5 m, because at t = t_0, my ball is up here at 2.5 m
again we can write for part (a), to find out what happens
as it gets to that point ...
it is good to write everything out the first time for completeness
but of course we know some of the stuff goes away such as a_x is equal to zero
y_0 is not equal to 0 this time, but x_0 is equal to 0
collecting our informational, we know ...
...
and we are given x_1, because it has to travel
all the way to the net
so 11.9 m, positive, because we have defined that way to be position
and y_1 isn't 0 either, it is at the top of the net, so 0.91 m, we have those
we don't know what t_1 is, but we do know what a_x and a_y is
and then we can also break down and work out my v_0x and v_0y 
but before we get to do that, they are pesky and gave us the speed in km/h
but everything else is in meters and probably seconds, so let's convert the v_0
into m/s
and this you have probably done so much that it is second nature now, through the calculator, you get ...
keeping a few extra digits, there you go
so this next bit is going to get a little messy
because we don't know theta, the only thing we can say ....
... and v_0y is going to be negative because the angle is below the horizontal ...
even though we know a number for v_0, we don't know the number for theta yet
so let's see what happens when we try to sub things in, x_1 we know ...
and then for y, we do a very similar thing
v_0y is negative, don't forget
so for both of these equations, we can't get anywhere because we have theta and t in it
and since we are interested in theta, let's eliminate t
by using the top [equation] and solving for t
because both these terms are 0, it's as simple as taking all this mess and dumping it underneath
...
...
...the units again work out, we get just s on top, and then we sub in t into these two spots
so that's t subbed in there
seconds cancels out, and this
number cancels out as well which is nice
let's collect that together, so ...
...
so we still getting m on both sides which is good
let's make it a little cleaner, there is only so much we can do here
we end up with a somewhat lengthy equation
that we can't really get much into
because it has got both sin(theta), and cos(theta) and a cos^2(theta)
so therefore, what we have to resolve to do is to use our graphing calculator
to graph a line, call that y_1 and put all this as y_2
graphing those 2 lines and finding where they intersect to find us our theta, which is our unknown
and while you may have your graphing calculator with you, I can't really show this on my screen 
so I instead use in this free online web site that has a graphing calculator on it
at desmos.com, fairly easy to use
you enter in your equation using
the various buttons down here, including your functions
and so I have already done that, I have got my left hand side: y = 0.91
and my right hand side where theta is replaced with x 
I get some really crazy graphs
but let's not forget we should be using degrees not radians
and so we can find out this particular point which is our intersect and that's 6.117°
from the graphing calculator we got that our theta is 6.11°
so that answers our part (a)
for complete a sake
the serve will have to be 6.1
now that the ball has clear the net, the second question in part (b) is
does it land in or not
so say we have tennis player here again
serving, we are basically trying to find out how far it lands
since we set our x = 0 way back here, let's just solve for what x_2 is first
this being t at t_2, again we don't know what time it is
but we already have our theta, which is going to help us quite a bit in our calculation
we are going to relate all the way back to the beginning, so we will use x_0 ...
...
and then we have y_2, which we know is 0 because it is at the ground now
looking at up top
we don't have x_2 and we don't have t
so let's deal with the other equation first
here we have ...
...
...
...
...
...
...
whoops, I forgot a time here
and now we're having another little bit of a tougher algebraic problem 
because we have t and t^2 in the same equation
well, to solve that, we need to employ the quadratic formula
reminder of what the quadratic formula is
I will use t in terms of the unknown instead of x as you may be used to 
because t is our unknown in this case
as long as we have this form: something times t^2 plus something times t plus something is equal to 0
this equal to zero very very important
then we have that t is equal to ...
...
this is something you might want to
keep handy and available in terms of your formula sheets
in my case, a is equal to this chunk ...
we will drop the units for that because we can see and check that
t will have to be in s for it to all work out
b is this whole thing ...
...
...
putting it into here, doing some calculator work, you will find two answers ...
...
in our case, of course we expect time to go forward
from the serve to the landing, so we will reject the negative and we have our time
with our time, we sub it into my x kinematics equation for position
and we can get what we need
...
...
...
...
.... well what does that mean?
thinking back to the diagram
what we have found is that from 
here it clears the net, to the landing spot is 17.195 m
but that's not what we are ultimately interested in
we just want to make sure that bit is less than 6.4 m
in front of the net
then to find that out, of course, we have to subtract out the original
11.9 m
so this particular distance here
...
...
5.3 m which is in fact to less than the 6.4 that is required, so 
the ball 
lands in
and it is a valid serve
definitely lengthier of a question
and uses some more advanced mathematical tools 
so that's why I wanted to show you this particular question
included the graphing calculator and also
the quartet formula
