>> In this video,
we're going to take
a closer look at
Advanced Projectile Motion
problems using
kinematic equations.
Our first problem states
that a ball is launched
at an angle of 45 degrees
above the ground
at 16 meters per second.
What is the maximum height
the ball reaches?
We start by making a note
of the given variables
and what we're
solving for.
The ball's initial velocity
is 16 meters per second,
so we write that down.
And the ball is launched
at an angle of 45 degrees.
Part of that
initial velocity
lies in the vertical
"y" direction
and part of it is in
the horizontal "x" direction.
We're going to figure out
the maximum height
the ball reaches--
y sub max.
Let's think about
what kinematic equation
we can start with that fits
the variables we're given.
We are solving for
a variable in the y direction.
Gravity is the only force
acting on the ball,
and is causing it
to accelerate and decelerate
in this direction.
We can use one of our four
kinematic equations
to help us solve.
The second equation has
an initial and final velocity,
a constant acceleration,
and a vertical displacement, y.
Since we want to solve
for the maximum height,
we know that the ball's
vertical velocity
at that point is zero.
That's the final velocity.
The ball's height is
the vertical displacement
from the ground.
So this equation
will let us solve
for the maximum height
of the ball.
The final vertical velocity
of the ball
when it's at its highest point
is zero.
The initial velocity
of the ball in the y direction,
v sub yi, equals
the initial velocity of the ball
times the sine of the angle
at launch.
We find this by taking
16 meters per second
times the sine of 45 degrees,
which makes our initial
velocity in the y direction
11.3 meters per second.
Now we arrange
our kinematic equation
to solve for y sub max.
We know our final velocity
in the y direction
is equal to zero,
so let's make that term
zero in our equation.
Subtracting v sub yi squared,
dividing both sides by 2a
leaves us with y sub max
equal to negative the initial
velocity in the y direction
squared divided by two
times the acceleration
due to gravity, a.
Plugging our known values
into the equation,
we find that the maximum height
the ball reaches
is 6.5 meters.
Next problem involves
solving for the position
and velocity of a projectile
after it has traveled
for several seconds.
As always with
projectile motion problems,
we write down
what we're given
and what to solve for.
We know the initial speed
of the baseball--
37 meters per second.
We know it left the bat
at an angle of 47.8 degrees.
We want to know the ball's
position and velocity
after t equal to 3.5 seconds.
Let's solve for the position
first in the x direction
and then in the y direction.
We consider only the motion
in the x direction.
We know that our ball
is not accelerating
in the x direction,
since no force is acting on it
in this direction.
The only equation
we can use in the x direction
is our constant velocity
equation-- v equals x over t.
Our velocity in the x direction
is always the same
along our journey.
Since the ball doesn't
accelerate and moves
at a constant velocity
in the x direction,
it is equal to the initial
velocity, which is
37 meters per second, times
the cosine of 47.8 degrees,
equals 25 meters per second.
Time is 3.5 seconds,
and we're solving for
the horizontal position
of the ball, x,
at this time.
When we rearrange
our equation, we find that
the x position of the ball
equals the velocity
in the x direction
times the time.
The x position of the ball
at 3.5 seconds equals
the velocity in the x direction,
25 meters per second,
times 3.5 seconds.
This equals 87.5
or 88 meters.
The ball's vertical position
at time 3.5 seconds--
we use one of our
acceleration equations.
Of the four, which one
should we use?
It looks like we have
enough information to use
the third equation.
Y equals initial velocity
in the y direction times time
plus 1/2 acceleration
times time squared.
Look at what we know.
Our initial velocity
in the y direction is equal
to the initial velocity
of 37 meters per second
times the sine of 47.8 degrees,
which equals
27.4 meters per second.
Time is 3.5 seconds.
Acceleration is negative
9.8 meters per second squared.
When we plug our values
into our equation,
we get y equals
27.4 meters per second
times 3.5 seconds
plus 1/2 times
negative 9.8 meters
per second squared
times 3.5 seconds squared,
equals 36 meters.
The position of the baseball
at time equals 3.5 seconds
is 88 meters
in the x direction
and 36 meters
in the y direction.
Now, what about the ball's
velocity at 3.5 seconds?
To find it, we need to
solve for the y velocity
after 3.5 seconds--
v sub yf--
and the final velocity
in the x direction-- v sub x--
and use the Pythagorean
theorem to find
the final resultant velocity
of the ball.
To solve for y velocity,
we will use the
kinematic equation that says
final velocity equals
initial velocity
plus acceleration times time.
Already determined that
the initial velocity of the ball
in the direction is
27.4 meters per second.
We have our time--
3.5 seconds--
and acceleration-- negative
9.8 meters per second squared.
We plug all of these values
into our equation
to find our final velocity
at this time in the y direction
is negative
6.9 meters per second.
Finally, we'll find
the x velocity of the baseball.
We know the ball's velocity
in the x direction is constant.
It will always equal
the same number
as long as the ball
is in the air.
It is the same number we got
for x direction velocity
as before--
37 meters per second
times the cosine
of 47.8 degrees.
That equals
25 meters per second.
Now we use these two values
as legs of our triangles,
and use the Pythagorean theorem
in order to solve
for the resultant,
our final velocity.
Final velocity is equal to
the square root of
the final velocity
in the x direction squared
plus the final velocity
in the y direction squared.
When we plug in our values,
we see the final velocity
is equal to the square root
of 25 meters per second squared,
which is 625 meters squared
per second squared,
plus negative
6.9 meters per second squared,
which is 47.6 meters squared
per second squared.
After we add these together,
we find the final velocity
is now equal to
the square root of
672.6 meters squared
per second squared.
When we take this square root,
our final answer is
25.9 meters per second.
You try this final
projectile problem
and see if you can solve for
the range the ball travels.
To begin, let's write down
the information given
and what we're solving for.
The ball's initial direction
is 25 degrees above horizontal.
Its initial speed is
30 meters per second.
The cliff's height--
we'll call it y sub cliff--
is 120 meters.
Solving for the overall range
the ball travels, x.
To know how far
the ball travels, we need to
first figure out how long
the ball is in the air.
The total time the ball
in the air-- t sub total--
equals the time it is
above its launch elevation--
t sub 1-- plus the time it is
below that point-- t sub 2.
First, let's solve for
t sub 1.
If we're ignoring
air resistance, we know that
the ball's vertical velocity
when it descends back to
the height it was launched from
is equal and opposite
its initial vertical velocity.
Since we're given
the initial velocity,
we can solve for
the final velocity.
Knowing initial and final
velocity and acceleration
and wanting to find t,
which kinematic equation
should we use?
The first one-- v sub yf
equals v sub yi
plus a times t--
is the best one.
If we subtract v sub yi
from both sides
and divide by
the acceleration, a,
we see that time equals
final velocity
minus initial velocity
divided by a.
The initial y velocity
is 30 meters per second
times the sine of 25 degrees.
That equals
13 meters per second.
V sub f-- the vertical velocity
of the ball when it returns
to its starting height--
is negative
13 meters per second.
It is traveling
in the opposite direction.
A-- the acceleration--
equals negative
9.8 meters per second squared.
Plugging all these numbers in,
we find that t sub 1
equals 2.7 seconds.
To solve for t sub 2,
we use the third
kinematic equation that says
displacement equals v sub yi
times t plus 1/2
a times t squared.
We know the displacement--
y sub cliff--
as negative 120 meters.
It is negative because
it is away from the origin
in the negative direction.
The initial velocity--
v sub yi-- that's
negative 13 meters per second.
And the acceleration-- a--
which is still a negative
9.8 meters per second squared,
since we're now speeding up
in the negative direction.
We want to solve for t2.
We add 120 meters
to both sides.
Notice that this equation
is now in the form of
a quadratic equation.
It is an equation in the form
zero equals a times x squared
plus b times x
plus c.
The two roots of a quadratic
equation are given
by the formula
x equals negative b
plus or minus the square root
of b squared
minus 4 times a
times c
divided by 2 times a.
In our case,
comparing our equation,
solving for t sub 2
with the equation
solving for x, we see that
a equals 1/2 times g--
the acceleration
due to gravity--
which all together is negative
4.9 meters per second squared.
B equals negative
13 meters per second,
and c equals 120 meters.
Plugging our values for
a, b, and c into the equation,
we find that the two roots
of t2 are 3.8 seconds
and negative 6.4 seconds.
We know which root to choose--
the positive one,
because time cannot be
negative.
Looking again at our diagram,
the total time
the ball is in the air
is t1 plus t2,
or 2.7 seconds
plus 3.8 seconds,
which, added together,
equals 6.5 seconds.
The range the ball traveled
equals its horizontal velocity--
v sub x-- times the total time
it is in the air.
We know t total,
but need to solve
for the velocity of the ball
in the x direction.
V sub x equals the initial
velocity-- v sub i--
times the cosine of the
initial launch angle-- theta.
V sub i is 30 meters per second
and theta is 25 degrees.
Plugging those numbers in,
we find v sub x
is 27 meters per second.
So the range--
the total horizontal distance
the ball traveled
from launch to landing--
is 27 meters per second
times 6.5 seconds,
which is 175.5 meters, which
to two significant figures
equals 180 meters.
That's a closer look at
advanced projectile motion
problems using
kinematic equations.
For more "Physics in Motion"
go to our homepage,
where the entire series
is available to you.