Video Transcript
In this video, we’re going to learn
about the acceleration due to gravity. What it is, the physical effects
that it has, and how to include this acceleration as we work through practice
problems.
As we start out, imagine that you
are competing in an egg drop competition. This is a competition where
competitors will wrap an egg in padding then drop their egg with its padding from
various heights and see who can drop their egg from the highest height without the
egg breaking. After long hours of preparation,
you have devised a support system for your egg which you hope will allow you to drop
it from the highest height and win the competition.
After careful testing, you found
that your egg support system can reach a maximum speed before impact after which the
egg remains intact and doesn’t break of 8.0 meters per second. The question is, given this maximum
speed, what’s the highest height you can drop your egg from without the egg
breaking? To answer that question, we’ll need
to know about the acceleration due to gravity.
To better understand the
acceleration due to gravity, let’s step back and consider the Earth as a gigantic
mass that we’ll call 𝑚. When a massive object exists, it
creates a field around it. This field draws other massive
objects in toward it. Through careful measurements, we
found out that the closer we are to the mass creating a field, the more we feel
pulled towards that mass, whereas the farther away we are, the less we feel
attracted.
If we were to take a mass, call it
capital 𝑀, and put it on the surface of the Earth, then the effect of the Earth’s
attractive field on that mass, capital 𝑀, can be quantified by something called the
object’s weight, symbolized 𝑤. The weight of an object on the
surface of the Earth is equal to the product of the mass of that object with the
acceleration, or the attraction, created by the mass of Earth on that object.
Through experimentation, it’s been
determined what that attraction due to the Earth is near the Earth’s surface. As a number, the value of that
attraction is about 9.8 meters per second squared. That’s how much the Earth will
accelerate a mass of object on its surface. Now, we may be thinking that makes
sense for an object which is in free fall, that is falling towards the Earth. But what about one like our mass
capital 𝑀 that rests on the surface of the Earth? How can we say that’s affected by
an acceleration when clearly it’s not moving?
In the case of our mass capital 𝑀
or any mass that rest on the surface of the Earth, another reactive force helps to
keep it from falling towards the center of the Earth. But still the mass is within the
gravitational field of the Earth and, therefore, experiences this acceleration of
about 9.8 meters per second squared. This number by the way is referred
to in shorthand as lowercase 𝑔. And it’s called the acceleration
due to gravity.
If we insert this symbol into our
equation, so it now reads 𝑤 equals 𝑀 times 𝑔, this equation may you look familiar
to something we’ve seen elsewhere. Newton’s second law of motion says
that the net force on an object is equal to the object’s mass times its acceleration
𝑎. Comparing these two equations, we
see that the two masses line up with one another. And weight is a force and can be
represented by the symbol 𝑤 sub 𝐹. So, this shows us more clearly that
𝑔, which is a quantity due to the Earth’s gravitational attraction, is an
acceleration, that is, a quantity with units of meters per second squared.
So, the Earth has mass and,
therefore, creates a field around itself that attracts other masses. And near the surface of the Earth,
when we measure this attraction, we find it’s approximately 9.8 meters per second
squared. What does that mean practically for
an object that we drop from a certain height? Let’s say we walk up to the edge of
a steep cliff and drop a rock so that it enters into free fall. We can imagine further that we have
a stopwatch where we keep track of the time after the rock’s release. And we want to understand its
position and its speed at each of these time increments.
Since we release the rock from
rest, we know its initial speed will be zero meters per second. As soon as we release that rock, it
will begin to accelerate at a rate of 9.8 meters per second squared, which means it
will speed up 9.8 meters per second every second that it falls. This means that at 𝑡 equals one
second, the rock’s velocity will be 9.8 meters per second. And at 𝑡 equals two seconds, its
velocity will be 9.8 meters per second plus 9.8 meters per second, or 19.6 meters
per second. And at 𝑡 equals three seconds,
we’ll add 9.8 meters per second to that. And the rock will now be speeding
along at 29.4 meters per second. That’s how the acceleration due to
gravity affects the speed of a descending object over time.
We can also use 𝑔, the
acceleration due to gravity, to understand the distances covered by the rock
vertically over these one-second increments. From when we first drop it to a
time of one second, the rock will fall 4.9 meters. As the rock gains speed due to the
acceleration caused by gravity, it covers more ground per second. And from zero to two seconds, it
covers 19.6 meters of vertical distance. And if we wait till three seconds
after dropping the rock, it will be 44.1 meters below the point where we released
it.
This is how the acceleration due to
gravity 𝑔 affects objects in free fall. The numbers we’ve calculated in
this example, by the way, come from using the kinematic equations. Let’s get some practice using those
equations with acceleration caused by gravity.
A coin is dropped from a hot-air
balloon that is 356 meters above the ground and rising at 12.3 meters per second
vertically upward. Assume that vertically upward
displacement corresponds to positive values. Find the distance between the coin
and the ground 3.07 seconds after being released. Find the velocity of the coin 3.07
seconds after being released. Find the time before the coin hits
the ground.
In this multipart exercise, first,
we wanna solve for the distance between the coin and the ground 3.07 seconds after
the coin is released. We’ll call that distance 𝑑. We also wanna solve for the
velocity of the coin that same amount of time after it’s released. We’ll call this velocity 𝑣. And finally, we want to solve for
the time it takes before the coin hits the ground. We’ll label this time 𝑡.
We’re told that at the moment the
coin is released, the hot-air balloon it’s released from is 356 meters above ground
and rising at 12.3 meters per second. Let’s begin our solution by drawing
a sketch of this scenario. We’re told that our hot-air balloon
is ascending at 12.3 meters per second in a direction upward, which we call the
positive direction. And during this ascent, a coin is
dropped from the balloon at a moment when the coin is 356 meters above ground
level.
Knowing all this, we want to solve
for the distance the coin falls. The distance 𝑑 is the distance
from ground level to the elevation of the coin after it’s been falling for 3.07
seconds. We also want to solve for the
velocity of the coin after it’s been falling for that same amount of time. And we want to find out the total
time it takes for the coin to reach the ground.
We know that when the coin is
released, it speeds up towards the Earth due to the acceleration caused by gravity
𝑔. Since we’ve established motion
vertically up as motion in the positive direction, we can write 𝑔 is equal to
negative 9.8 meters per second squared. That acceleration 𝑔 is the only
acceleration experienced by the coin. And it’s constant, which means that
the kinematic equations apply to this scenario.
These four equations, based on the
assumption that acceleration 𝑎 is constant, describe the motion of objects in free
fall, for example. Given that we want to solve for
distance 𝑑, that we know the coin’s initial velocity, we know the acceleration to
which the coin is subjected, and we also know the time after its release at which we
want to calculate 𝑑, this all points to using the third kinematic equation
listed.
We have to be a bit careful when
using this equation because it will tell us how to calculate a distance equal to the
distance that the coin descends over 3.07 seconds. But we, rather, want to solve for
the coin’s elevation after that amount of time. When we solve for lowercase 𝑑,
that will equal the total original elevation of the coin minus the distance it
descends over 3.07 seconds. We can write this as 356 meters
plus 𝑣 zero 𝑡 plus one-half 𝑎𝑡 squared because of the sign convention we’ve
chosen. We’ll see how this works out.
Looking at the three variables
involved in our expression, we know 𝑣 sub zero. That’s 12.3 meters per second, and
it’s positive. We also know 𝑡. We’re given that as 3.07 seconds as
well as 𝑎. That’s equal to negative 9.8 meters
per second squared. When we plug these values into our
expression and then enter this whole expression on our calculator, we find that
lowercase 𝑑 is equal to the original distance from the coin to the ground plus
negative 8.42 meters. In other words, over 3.07 seconds,
the coin descends 8.42 meters. That means that, rounded to three
significant figures, it’s 348 meters above the ground after 3.07 seconds of free
fall.
Next, we want to solve for the
coin’s velocity 3.07 seconds after it’s released. And again, we’ll look through our
list of kinematic equations to search for a match. The first equation involves
acceleration which is known, initial velocity which is known, final velocity which
we want to solve for, and time which is known.
When we write out this relationship
using our variables, we see that 𝑣 sub zero, the initial velocity of the coin, is
known because that’s the initial velocity of the hot-air balloon. We also know the acceleration and
𝑡. When we plug these three values
into our equation and calculate 𝑣, we find the result of negative 17.8 meters per
second. That’s the velocity of the coin
3.07 seconds after it began to fall.
Lastly, we want to solve for 𝑡 the
time it takes for the coin to reach the ground. Once again, we’ll use a kinematic
equation to help us solve for this value. The third kinematic equation
written includes the variable we want to solve for 𝑡 as well as acceleration which
we know and the coin’s initial velocity 𝑣 sub zero. The distance that the coin falls in
total, based on our diagram, is capital 𝐷 plus lowercase 𝑑.
Because this total displacement is
in the negative direction by our sign convention, we write that it’s negative 356
meters, a magnitude of distance given in our problem statement. If we take a next step by adding
356 meters to both sides of our equation, we see that we now have an equation in the
form of a quadratic equation, where the variable we want to solve for is the time
𝑡.
In our equation, the constant
factor in front of the squared term, which we can call capital 𝐴, is equal to
one-half times negative 9.8 meters per second squared. Our constant factor in front of the
linear term, which we can call 𝐵, is equal to positive 12.3 meters per second. And 𝐶 is equal to 356 meters. We can recall that the quadratic
equation tells us, if we have an equation of this form, then the roots of that
equation, represented here by 𝑋, are equal to negative 𝐵 plus or minus the square
root of 𝐵 squared minus four 𝐴𝐶 all divided by two 𝐴. Where we’ve identified capital 𝐴,
capital 𝐵, and capital 𝐶 for our equation.
Using our 𝐴, 𝐵, and 𝐶 values
and, in our case, solving for the roots of the time 𝑡, when we plug in our 𝐴, 𝐵,
and 𝐶 values into the quadratic equation to solve for 𝑡, we find that the two
roots are 9.87 seconds and negative 7.36 seconds. Since we know the time value for
the coin to drop to the ground can’t be negative, we eliminate that root and choose
the first one. So, 9.87 seconds is the amount of
time it takes for the coin to go from being released to hitting the ground.
Now, let’s take a moment to
summarize what we’ve learnt about acceleration due to gravity. Gravity has an effect on every
object in its field. And it accelerates objects which
are in free fall or released. The acceleration caused by gravity
is represented by the symbol lowercase 𝑔. Its magnitude is 9.8 meters per
second squared. And it points downward towards the
Earth. And finally, the kinematic
equations explain object motion when acceleration is equal to 𝑔, the acceleration
due to gravity.
