### Video Transcript

A kangaroo can jump over an object 2.50 meters high. What is the initial vertically upward speed of the kangaroo that jumps over the object? How much time is the kangaroo in the air for during its jump over the object?

Let’s begin this solution by pulling out some of the important information from the problem statement. So we’re told that the kangaroo is capable of jumping over an object that’s two and a half meters high, 2.50 meters. And what we like to find is the initial upward vertical speed of the kangaroo, and we also wanna find how much time the kangaroo takes to jump up over the object and then land back down on the ground.

We can represent this information using variables. So let’s choose ℎ to represent our height of 2.50 meters. Let’s choose 𝑣 sub 𝑣 for vertical speed to represent our initial vertically upward speed, that’s something we’re trying to solve for. And let’s let 𝑡 represent time, that’s again the time that the kangaroo is in the air, and that is also something we’re looking to solve for.

If we draw a picture of what’s going on, so we’ve got our kangaroo — looks a little bit like a mouse or a rat here but trust me, it’s a kangaroo — and the kangaroo is jumping up just clear as our two and a half meter tall object and then comes back down and lands. And what we would like to know is when it jumps, what is it’s initial vertical or upward speed and how long does this whole jump take while the kangaroo is in the air.

You may see this is a kinematics problem, so a list of our kinematics equations will help us figure out how to make headway with this problem. So here is our list of kinematic equations and like always, what we’ll do is we’ll scan the list and look for an equation that fits the information we’re given. So what we wanna solve for again is the vertical upward velocity of the kangaroo and then subsequently, the time that this kangaroo takes to complete it’s jump.

Now take a look at this second kinematic equation here. It says that the final velocity squared of our object equals the initial velocity squared plus two times its acceleration times the displacement or the distance that it travels. As we look at that equation, we find it’s a match for the information we’re given. Now you might say, “Well, how is it a match?” So let’s look more carefully at it.

So our first term, 𝑣 sub 𝑓, that’s the final velocity of the kangaroo in this case. Now let’s look a little bit at this diagram and focus here on this point, at the highest point that the kangaroo ever is, two and a half meters off the ground. Now, if we let that be our final point, then we know that the vertical velocity of the kangaroo at that point, at that 𝑥, is zero. Now that’s very helpful because it means this whole term will cancel. The velocity there is zero meters per second. Now this term 𝑣 sub 𝑖, we’ve represented that as our initial vertical speed of the kangaroo; we’ve called it 𝑣 sub 𝑣. Now moving along, our acceleration in this case is gravity which we represent by a lowercase 𝑔. And the 𝑑 in this equation, we’ve called ℎ; that’s our height of two and a half meters. So if we rewrite this kinematic equation in terms of our particular situation, we get that zero equals 𝑣 sub 𝑣, the initial upward speed of our kangaroo squared plus two times 𝑔, the acceleration due to gravity times ℎ, the maximum height that the kangaroo achieves.

Before we go any further, let’s define which direction in this case is positive. You might say it’s implicitly defined already. But just to be very clear, let’s say that up in our case is the positive direction. So any velocity or acceleration or motion in that direction is positive and anything downward, we’ll call negative.

Alright, now that we’ve gathered our information, we’ve picked the kinematic equation to use and applied it to our own situation with our variables, we’re ready to start solving for 𝑣 sub 𝑣, the initial upward speed of the kangaroo. Now taking a look at this equation here, first what we wanna do is rearrange this equation so that we can solve for 𝑣 sub 𝑣 in terms of our given variables. So let’s start that. Let’s start by subtracting two times 𝑔 times ℎ from both sides of the equation. You can see on the right hand, that term cancels out. And now, if we take the square root of both sides of the equation, the square and the square root on the right side once again cancel out, and we’re left with the equation that says 𝑣 sub 𝑣 is equal to the square root of negative two times 𝑔 times ℎ.

As we plug-in for these values, we see that 𝑔 is a negative 9.8 meters per second squared. Now you might say, “Well, why is it negative?” Well, remember that we’ve defined up as positive, so any motion, including acceleration, that points in the opposite direction, we’ll define as negative. And that gives us our negative sign here in this term. And then moving on, when we insert our value for ℎ, we know that that is 2.50 meters.

Now when we plug these values in to our calculator, we end up with a final initial upward velocity of the kangaroo of 7.00 meters per second. This result agrees with the number of significant figures we were given for ℎ, the height of the obstacle.

Now let’s move on to solving for 𝑡, the time the kangaroo is in the air. On our way to solving for 𝑡, the time that the kangaroo is in the air, we can make a shorthand that will help us in our solution. So see this-this 𝑥 here, the top of our object, now because we’re neglecting air resistance, the time it takes for the kangaroo to go from the ground up to the 𝑥 compared to the time it takes for the kangaroo to go from the 𝑥 to the ground is the same. So we can call the time elapsed when the kangaroo is at its upper highest position 𝑡 one-half, that’s half the total time that the kangaroo is in the air. So we can write that as 𝑡 one-half equals 𝑡 over two. That’s useful because it lets us take advantage of the fact that the velocity or the speed of the kangaroo at that point when 𝑡 equals 𝑡 one-half is zero.

Speaking of kinematic equations, let’s look again at our list of those equations and see if we find something useful. So here again is our list of kinematic equations and as we look at the very first equation, we see it looks promising. We have 𝑣 sub 𝑓, the final velocity, is equal to 𝑣 sub 𝑖, the initial velocity, plus acceleration times time. Now just like we did for the initial upward speed, let’s map this equation and these variables to our scenario.

Now if we look at 𝑣 sub 𝑓, remember again that we’re thinking of the time when the kangaroo is at its highest point. And at that point, the final velocity of the kangaroo is zero meters per second. Now the initial velocity of the kangaroo, 𝑣 sub 𝑖, that’s something we just solved for. That’s 𝑣 sub 𝑣 which equals 7.00 meters per second. Now the 𝑎 in this equation is equal to lowercase 𝑔, the acceleration due to gravity. And 𝑡, we’ve made a special notation and we’ve created a variable we’ve called 𝑡 sub one-half. That’s the time elapsed from the kangaroo leaving the ground to being at its maximum height.

So if we rewrite this kinematic equation in terms of our own variables, it’ll look like this: zero equals 𝑣 sub 𝑣 plus 𝑔 times 𝑡 one-half. And remember, what we’re driving at is to be able to solve for 𝑡 one-half. And then to take advantage of the fact that finding 𝑡 one-half will ultimately let us solve for 𝑡. So let’s begin doing that by rearranging this equation.

The first thing we can do to isolate 𝑡 sub one-half by itself is to subtract 𝑣 sub 𝑣 from both sides of our equation. What that will do is cancel that term out on the right-hand side of our equation. Now we’ll divide both sides of the equation by 𝑔, the acceleration due to gravity. And again, that cancels out on the right-hand side leaving us with an equation that says: 𝑡 sub one-half is equal to negative 𝑣 sub 𝑣 divided by 𝑔, the acceleration due to gravity. When we plug in these numbers, remember we have 7.00 meters per second as our initial upward speed and 𝑔 is negative 9.8 meters per second squared. And once again, it’s negative because our convention is to say that motion up is positive and therefore motion down, including acceleration, is negative.

Now when we plug these numbers in to our calculator, you see the negative signs cancel out and we get a result which says that 𝑡 one-half is equal to 0.714 seconds. But we’re not quite done yet. Remember that we wanna solve for 𝑡 rather than 𝑡 sub one-half. Now we know that 𝑡 is equal to two times 𝑡 sub one-half, so that lets us multiply two times 0.714 seconds.

And our final result is that 𝑡 is equal to 1.44 seconds. That’s the total time the kangaroo is in the air during its jump.