# Video: Finding the Work Done by Drag Forces on an Object in Ballistic Motion

A boy throws a ball of mass 0.25 kg vertically upward with an initial speed of 20 m/s. The ball moves vertically upward, comes instantaneously to rest, and then accelerates vertically downward and returns to the point from which it was thrown. When the ball returns to its original position, its speed is 17 m/s. How much work was done by air resistance on the ball during its flight?

04:23

### Video Transcript

A boy throws a ball of mass 0.25 kilograms vertically upward with an initial speed of 20 metres per second. The ball moves vertically upward, comes instantaneously to rest, and then accelerates vertically downward and returns to the point from which it was thrown. When the ball returns to its original position, its speed is 17 metres per second. How much work was done by air resistance on the ball during its flight?

Let’s start by highlighting some of the critical information we’ve been given. We’re told the mass of this ball of 0.25 kilograms and that initially it’s thrown upward with a speed of 20 metres per second. Then, we’re told that when the ball comes to rest and descends back to its start point — when it reaches that original position — its speed is now 17 metres per second. And we want to figure out how much work was done by air resistance on the ball during its flight. We can symbolise that work with capital 𝑊 sub 𝑎 for the work due to air resistance.

Now, let’s begin our solution by recalling a theorem that connects energy with work. This theorem is appropriately named the work energy theorem and it states that the work done on an object is equal to the change in kinetic energy of that object. We’ll make use of this theorem to find the work done by air resistance, 𝑊 sub 𝑎. Now, let’s begin to create some symbols for the information we’ve been given about our scenario.

We’re told the mass of the ball is 0.25 kilograms; we can call that lowercase 𝑚. We’re also told that the ball is initially thrown upward at a speed of 20 metres per second; let’s represent that as 𝑣 sub 𝑖, the initial speed of the ball. Finally, we’re told that when the ball returns to its original position, it has a new speed of 17 metres per second. We’ll represent that as 𝑣 sub 𝑓.

Now that all that is in place, let’s write out an application of the work energy theorem for our particular scenario. We know we’re going to see a change in kinetic energy because the initial speed and final speed of the ball are not the same. The reason for this loss of speed has to do with the work done by air resistance, slowing the ball down as it moves up and also as it comes down.

So 𝑊 sub 𝑎, the work done by air resistance, is equal to the change in the ball’s kinetic energy. That change in energy can be written as the initial kinetic energy of the ball minus its final kinetic energy. Now we’re at a good point to recall the equation for the kinetic energy of an object. The kinetic energy of some object is equal to half the mass of the object times its velocity squared.

Let’s rewrite our equation for the work done by air resistance in terms of those terms of mass and speed. We see that the work done by air resistance is equal to one-half the mass of the ball times its initial speed squared minus one-half the mass of the ball times its final speed squared. We see that both these terms have one-half the mass of the ball in common, so we can factor that out on the right side of our equation.

Now, we can plug in for our values of 𝑚 𝑣 sub 𝑖 and 𝑣 sub 𝑓 from the information given in the problem: 𝑚 is 0.25 kilograms, 𝑣 sub 𝑖 is twenty metres per second, and 𝑣 sub 𝑓 is 17 metres per second. With all these numbers plugged in, when we enter these terms on our calculator, we find that the result is 14 joules. That is the total work done by air resistance in this process, which we use the work energy theorem to help us solve.