in this video we have a projectile motion problem and it's an elevated launch position so this
cannonball is shot from 72 meters above sea level and then it lands all the way down at sea level
well that breaks the symmetry of the problem so there's a very common formula in the textbooks
that gives you the range of a projectile based on the launch angle and the speed but that formula
only applies to a level surface so that no longer applies we have to take a more general approach to
everything and in this video i've asked basically everything that i could ask about the projectile
so in part a i want the flight time in part b i want the range in part c i want the maximum height
above the c and in part d i want the impact speed when the projectile hits the water so the prep
work that we normally do before anything else is to break that launch velocity into its components
and the x component is going to be 150 cosine of 38 and that gives me about 118.2 meters per second
so the projectile is moving that fast to the right simultaneously it's moving up and that component
is given by 150 sine of 38 and that's about 92.3 meters per second one last thing i'll do as prep
work is just to very clearly label what my origin is and i want to call the initial x position 0
and then i'm going to make c level the origin of coordinates for the y coordinate
and in part a i'm asked for the flight time of the cannonball and because i know the final y
position when this thing lands i have a little bit of extra information about the vertical analysis
of the flight and that's how i know to start this question a by looking at the vertical analysis
so the y motion of the canon ball follows the standard kinematics equation y equals y naught
plus v naught t plus one half a y times t squared because the acceleration in the y direction is
constant and remember that acceleration is the acceleration induced by gravity so that'll be
negative 9.8 meters per second squared my final y position for the flight is zero my initial 72
meters off the surface of the ocean my initial y velocity was about 92.3 meters per second
and again i'm going to sub in a y equals negative 9.8 half of that is 4.9 so i get negative 4.9
t squared so we're left with a quadratic equation to solve and i prefer to have a positive
coefficient on the quadratic term so i'm going to move all of the terms to the left-hand side
and i've gotten to the point where i don't ask students to show me all the steps in the quadratic
formula anymore because so many students have an algorithm built into their calculator anyway
but if you want to show everything by hand you would use the quadratic formula at this point
and that starts with a negative b so that's the coefficient of
t negative b plus or minus the square root of b squared minus 4
times a times c and there's a double negative in that second term because c is negative
and then all divided by 2 times a and 2 times 4.9 is 9.8
so when you run the numbers on that square root piece
that simplifies to 99.65 and i can immediately tell i'm going to get a positive solution and a
negative solution and it's definitely the positive one that i want to keep
because our time starts at t equals zero it's interesting to wonder why there's a negative
solution in there and what that's about is just that the mathematical model we've built
doesn't include the physics of the launch itself it's just a model of an object in free fall moving
in two dimensions so our extraneous solution corresponds to the moment where the projectile was
flying up from sea level before it reached the tip of the canon at t equals zero well we know that
wasn't really happening physically so we discard that solution i'm going to keep only the plus in
here because it gives me a positive solution and that positive solution is 19.6 seconds
now in practice i don't see anything wrong with just relying on technology for using the quadratic
formula with a bunch of decimals so here we are in wx maxima that's a free computer algebra system
and i put in solve this equation so i just wrote down the equation for the variable t and then i
put the word float out in front which means give me a decimal approximation and i hit shift enter
and i can see my two solutions i have a negative 0.75 seconds or a positive 19.6 seconds that
of course is the one that we're going to keep the syntax would be very similar if you wanted to use
wolfram alpha for example in part b we're asked for the range of the cannonball
in other words what's the horizontal position relative to the launch position
when this thing lands and we normally use an r for that so i'll just pop that into the diagram so we
look at the horizontal analysis of the flight and i have x equals x naught plus v naught x t
just the standard equation of motion for constant acceleration kinematics
plus one half ax t squared and then a key feature of projectile motion is the
acceleration in the horizontal direction is zero in other words the x velocity is constant
so that acceleration term is gone and then we put the origin at the initial position
of the object so the initial x value is zero and i basically just get distance
equals rate times time out of this so my final x position i'll call that capital r
it's going to be my x velocity times the flight time
and when i crunch the numbers i get 2 320
meters just keeping three sig figs next i want to find the maximum height above the c
for this projectile and just a word of caution here you may be used to using half the flight
time in order to determine the maximum height but that only applies to a level surface so if i was
launching something from the same level that it lands on then it reaches its highest point
at half the flight time but again our symmetry is broken so we have to take a more fundamental
approach and the key physical fact for finding the maximum height is that it's the transition
between the object moving up and the object moving down at the transition between positive
and negative velocity the velocity is momentarily zero so v y is equal to zero at the maximum height
now the easiest way to get it to maximum height is to use our third kinematics formula
so provided we have constant acceleration in the y direction this formula is valid
and then i plug in the fact that the y velocity is zero
at the maximum height and then my initial y velocity that was 92.3
a in the y direction is negative 9.8 so i move the minus sign out in front there the y value
at this moment of interest that's what we're after so that's y and then my initial y value
was 72. so when i crunch the numbers on this to solve for y i get a maximum height of 507 meters
note that it's perfectly fine if you want to plug in v y equals zero into the kinematics equation
v equals v naught plus a t and then you could solve for the time of the maximum height and
then plug it in to this equation of motion for the y coordinate and you'll find the same value of 507
meters in part d we're asked for the impact speed when the ball hits the water
and this has to be done by components there is a sneaky way to do this using the concept of
gravitational potential energy but we don't have that yet at this point in the course
so i'm going to get into the details of the velocity components when the cannonball strikes
the surface of the water just to visualize that real quick there's the impact velocity vector
and that's made of a horizontal component that's v x final
and a vertical component v y final v x final is the easy one
because the x velocity never changes so i know immediately that that's 118.2 meters per second
v y final requires a bit of calculation again v y is v not y plus a y times t and
my initial y velocity was 92.3 meters per second my acceleration negative 9.8 my flight time 19.6
and i get a final y velocity of negative 99.8 meters per second the minus sign is
there because my final y velocity is downward then i have to vector add these two components
my x component is 118.2 my y component i'm just going to write the magnitude of it on
there 99.8 and the magnitude of my final velocity vector i'm just going to call v now it's just the
length of that that's equal to the speed when this thing hits the water i wasn't asked for the angle
so we just use the pythagorean theorem on this right triangle
and when i run the numbers i get an impact speed of 155 meters per second note that in some
problems you'll be asked for the angle at which the projectile strikes the ground and in that case
you could just quickly use an inverse tangent to find the size of that angle that i labeled
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