in this video we're given a projectile launched on a level surface with a known velocity
and we're trying to find the flight time and the range and the maximum height for the projectile
so the way we start things out is by just breaking that initial velocity into components which is
actually part a of the question so i normally do as much as i can up here in the diagram
and i'm going to break that velocity vector into a horizontal piece that i could call v
x and a vertical piece that i call v y and that horizontal piece is given by the adjacent leg
of the right triangle when we break this vector into components so that's given by the hypotenuse
45 meters per second multiplied by the cosine of the angle cosine 38. when i run the numbers
on this i get 35.5 meters per second for the horizontal component of velocity now the v y
we're looking at is actually the opposite leg of the right triangle the vertical component
of that original velocity vector but it's been translated to the initial position
of the object which you're free to do with vectors they're totally movable
and the magnitude then is the same as the opposite leg of that right triangle so 45 sine 38 and when
i run the numbers on this i get 27.7 meters per second so there's the answers to part a
in part b i want the flight time and the key to deciding how to approach this
is looking at the landing of the object and asking what extra bit of information do i have there
and for projectile motion problems that's either going to be a known final x-coordinate or a known
final y-coordinate and in this case it's a known final y-coordinate of zero so i'm going
to approach this by using the y analysis of the problem so the y coordinate as a function of time
here is given by the first kinematics equation and for projectile motion the acceleration in
the vertical direction is downward and has a magnitude of g or 9.8 meters per second squared
i'm going to go ahead and plug in my initial y value that was zero and my final y value
that's also zero and my initial y velocity that's 27.7 meters per second
and my acceleration that's negative 9.8 so i'm going to pull the minus sign out in front
and then half of 9.8 is 4.9 t squared now this is a quadratic equation but we don't
require the quadratic formula and i can solve this quadratic equation simply by factoring
i pull a t out and there's two solutions to this equation the easy one is that t is equal
to zero solves this equation and the harder one is found by setting this other factor equal to zero
so when that factor is equal to zero the equation is also true there because zero times anything is
zero so setting 27.7 minus 4.9 t equal to zero i can then add 4.9 t to both sides and then divide
both sides by 4.9 and i get t so that's 27.7 over 4.9 and i get that t is equal to 5.65
seconds now we got two answers out of this but we only care about one of them and to understand why
the extraneous solution was produced i go back to what i was telling the mathematics
here i was really just telling the equations y is equal to zero tell me what time it is and
there's two different times where this is true it's the starting point and the finishing point
now clearly it's the finishing point we're interested in so t equals zero
is our extraneous solution and our flight time is 5.65 seconds once you have the flight time
you can get the range very quickly i just look at the x equation of motion for the projectile
again that's just the first kinematics equation but then i plug in this information
about projectile motion there's no forces acting horizontally so the x acceleration is always zero
for projectile motion and this first kinematics equation reduces to basically
distance equals rate times time i'm going to go ahead and plug in the fact
that my origin was at the launch position as well and now it really looks like distance equals rate
times time so i want to get the final x position and i would normally call that r
and that's going to be the initial x velocity which was 35 and a half meters per second
times the flight time that's how long it's been moving rightward that fast 5.65 seconds
and rounding to three sig figs i get 201 meters for this finally i want to get the maximum
height for the projectile and there's two basic approaches to this because of the symmetry of the
problem because the landing height is the same as the launch height i can say this projectile
reaches its maximum height at half the flight time exactly just keep in mind that approach is
not going to work unless you have this high degree of symmetry if the landing height was higher or
lower than the launch height the maximum height would no longer occur at half the flight time
so rather than show that method i'm going to show the more general method which is to say that at
the maximum height that's the transition between a positive v y and a negative v y and right
at that transition point you're crossing over from a positive number to a negative number v
y must be zero at the maximum height and this is the general approach to getting maximum heights
in free fall problems and projectile motion problems you say v y equals zero and then you
ask what's the height at that point in time again we have options here i could use the fact that v
y is equal to zero and find out what time it is and then plug that time into the height equation
or i could get more clever about it and look for another kinematics formula that doesn't involve
time and that's this one again these formulas are all valid as long as the acceleration is
constant which it is in the y direction for projectile motion is constantly negative 9.8
all right so this offers me a bit of a shortcut i can plug in
the v y is equal to zero and i'm going to ask
what's the y coordinate at that moment in time my initial y coordinate also zero
so plugging in the numbers here i get zero equals v not y squared that's 27.7
squared minus because i'm going to plug in a equals negative 9.8
2 times 9.8 times my final y coordinate and now i just have to isolate that y so i'm going to
add this term to both sides of the equation and then divide by twice 9.8 so i type in 27.7 squared
divided by parentheses twice 9.8 and rounding to three sig figs i get a y value of 39.1
meters just to be extra thorough here i'll go ahead and use the half the flight time method
as well and just make sure that it that the answer agrees and again this only works because i have a
high degree of symmetry in this problem what i'm saying is because i have this perfectly symmetric
trajectory the highest point should happen at half the flight time and i'm going to call that t max
half the flight time is 2.83 seconds and then i plug into the vertical equation of motion and i
find out what's the y coordinate when t is equal to 2.83 seconds again my initial y value is zero
and my acceleration is negative 9.8 meters per second squared so i plug in
v naught that's 27.7 in the y direction times t minus one half times 9.8
times t squared 2.83 squared and when i run the numbers on this i get 39.1 meters just giving
us a double check on our work if you find the physics content on zack's lab helpful click on the
zaxlab logo on the right to browse playlists and subscribe to the channel i produce over 100 new
videos per month and subscribing is the easiest way to find new content thanks for watching