Question Video: Kinematics of a Falling Object

A coin is dropped from a hot-air balloon that is 356 m above the ground and rising at 12.3 m/s vertically upward. Assume that vertically upward displacement corresponds to positive values. Find the distance between the coin and the ground 3.07 s after being released. Find the velocity of the coin 3.07 s after being released. Find the time before the coin hits the ground.

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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.

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