Video Transcript
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.
