# Question Video: Finding the Difference in the Displacement of Two Objects Physics • 9th Grade

Two aircrafts scramble from the same airfield, one climbing at an angle of 45° from the ground and the other at 20° from the ground, as shown in the diagram. When both aircraft are horizontally 2500 m from the airfield, the aircraft that climbed at a less steep angle is vertically ℎ₁ meters above the ground and the aircraft that climbed more steeply is vertically ℎ₂ meters above the ground. How far vertically above ℎ₁ is ℎ₂, to the nearest meter? How much greater is the displacement from the airfield of the fast-climbing aircraft than the displacement from the airfield of the slow-climbing aircraft when the fast-climbing is ℎ₂ meters above the ground? Answer to the nearest meter.

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### Video Transcript

Two aircrafts scramble from the same airfield, one climbing at an angle of 45 degrees from the ground and the other at 20 degrees from the ground as shown in the diagram. When both aircraft are horizontally 2500 meters from the airfield, the aircraft that climbed at a less steep angle is vertically ℎ one meters above the ground and the aircraft that climbed more steeply is vertically ℎ two meters above the ground. How far vertically above ℎ one is ℎ two to the nearest meter? How much greater is the displacement from the airfield of the fast-climbing aircraft than the displacement from the airfield of the slow-climbing aircraft when the fast-climbing is ℎ two meters above the ground? Answer to the nearest meter.

Okay, so we’re told in this question that we’ve got two aircraft, each climbing at a different angle relative to the ground as it leaves an airfield. This is shown in the diagram where we can see that we’ve got one aircraft at an angle of 45 degrees to the ground and the other at an angle of 20 degrees. Also shown in the diagram is that when both aircrafts have traveled a distance of 2500 metres horizontally, the steeper-climbing aircraft has reached a height of ℎ two meters above the ground, and the less steep one has reached a height of ℎ one meters.

So, all of the information given to us in the main bit of question text is also shown in the diagram. This means that we can safely get rid of this main bit of text to clear ourselves some space. Let’s make a start with the first part of the question. This says, how far vertically above ℎ one is ℎ two to the nearest meter? We know that ℎ two is the height that the steeper-climbing aircraft has reached after traveling a horizontal distance of 2500 meters from the airfield. And we know that ℎ one is the height that the less steeply climbing aircraft has reached after traveling this same horizontal distance.

We’re asked how far vertically ℎ two is above ℎ one. So that’s this distance here in our diagram, and it must be equal to this distance here, which is ℎ two meters, minus this distance here, which is ℎ one meters. In other words, the question is asking us to calculate the value of ℎ two minus ℎ one. This means we need to start by calculating the values of ℎ two and ℎ one. Let’s begin with the quantity ℎ two, the height of the steeper-climbing aircraft.

We can identify a right-angled triangle in the diagram; the hypotenuse of this triangle is the path that the steeper-climbing aircraft moves along. And then there’s a horizontal side with a length of 2500 meters — that’s the horizontal distance moved by the aircraft — and a vertical side with a length of ℎ two. That’s the height that this aircraft has climbed when it’s traveled this 2500 meters horizontally. We know that the angle that the steeper-climbing aircraft makes to the ground is 45 degrees. And so that’s the value of this angle in the bottom-left corner of the triangle.

To find the value of ℎ two, we need to recall a useful trigonometric equation. Let’s consider a general right-angled triangle and suppose that this angle has some value of 𝜃. We’ll label the length of the side opposite this angle as 𝑜 for opposite and the length of the side adjacent to it as 𝑎 for adjacent. Then, for this general right-angled triangle, the trigonometric equation that we’re going to find useful says that tan 𝜃 is equal to 𝑜 divided by 𝑎. If we compare this general right-angled triangle with the one that we’ve identified in our diagram, then we can see precisely why this equation is going to be useful to us.

The equation connects the angle 𝜃, the length of the side adjacent to the angle, and the length of the side opposite it. That means that if we know any two of these quantities, we can use this equation to calculate the third. In the triangle in this diagram, we know the value of one of the angles. It’s 45 degrees. And so, that’s our value for the quantity 𝜃. We also know that the length of the side of the triangle adjacent to this angle is 2500 meters. So that’s our value of the adjacent side 𝑎.

This vertical side of the triangle has a length of ℎ two. And that’s what we’re trying to find here. This is the side of the triangle that’s opposite the angle 𝜃. And so, ℎ two is our value for the quantity 𝑜. So, we’ve got values for both 𝜃 and 𝑎, and we want to find the value of 𝑜. This means that we need to rearrange this equation to make 𝑜 the subject. To do this, we simply multiply both sides of the equation by 𝑎. Then, on the right-hand side, the 𝑎 in the numerator cancels with the 𝑎 in the denominator. And writing the equation the other way around, we have that 𝑜 is equal to 𝑎 multiplied by tan 𝜃.

Let’s now take our values for 𝑎 and 𝜃 and substitute them into the right-hand side of this equation. In place of 𝑜, we’ve written ℎ two because that’s the opposite side in this triangle and this is equal to 2500 meters, which is our value for 𝑎, multiplied by the tan of 45 degrees, which is our angle 𝜃. The tan of 45 degrees works out as simply one. And so, ℎ two is equal to 2500 meters multiplied by one, which is just 2500 meters. So, we’ve found our value for ℎ two, the height reached by the steeper-climbing aircraft.

We can notice that this height is the same as the horizontal distance of 2500 meters that the aircraft travels. Since this aircraft travels at an angle of 45 degrees to the ground, which is halfway between traveling horizontally and traveling vertically, then it makes sense that the vertical distance climbed is equal to the horizontal distance moved. Now that we’ve worked out the value of ℎ two, it’s time to move on to finding the value of ℎ one.

To do this, we need to identify a second right-angled triangle in our diagram; specifically, that’s this triangle shown in pink. The hypotenuse of this triangle is the path of the less steeply climbing aircraft. The bottom-left angle in this triangle is 20 degrees since that’s the angle that this aircraft makes to the ground. Both these two aircrafts travel the same horizontal distance. And so, this triangle has a horizontal side of length 2500 meters, just like the first one did. The vertical side of this triangle is the height ℎ one that this less steeply climbing aircraft reaches.

In the same way as we did before, we can identify our angle 𝜃 as 20 degrees, our adjacent side 𝑎 as 2500 meters, and our opposite side 𝑜 as ℎ one. Again, like we did before, we can take our values for 𝜃 and 𝑎 and substitute them into this equation. When we do this, we get that our opposite side ℎ one is equal to 2500 meters — that’s our adjacent side 𝑎 — multiplied by the tan of 20 degrees, which is our angle 𝜃. The tan of 20 degrees works out as 0.36397 and so on with further decimal places. And then evaluating this expression, we find that ℎ one is equal 909.9256 meters, again with further decimal places.

Now that we’ve worked out both ℎ two and ℎ one, then all we need to do to calculate this distance here, which is the vertical height of ℎ two above ℎ one, is to subtract our value of ℎ one from our value of ℎ two. Since we’ve seen that this distance is equal to ℎ two minus ℎ one. We have that ℎ two minus ℎ one is equal to 2500 meters minus 909.9256 meters, which works out as 1590.074 meters. We’re told to give our answer to the nearest meter. And so, rounding our result to the nearest meter gives us our answer to this first part of the question as 1590 meters.

Now, let’s clear some space so that we can look at the second part of the question.

How much greater is the displacement from the airfield of the fast-climbing aircraft from the displacement from the airfield of the slow-climbing aircraft when the fast-climbing is ℎ two meters above the ground? Answer to the nearest meter.

The displacement of each aircraft from the airfield is the straight-line distance from the airfield to that aircraft’s current position. So, when the fast-climbing aircraft is ℎ two meters above the ground, which is what’s depicted in the diagram, then the displacement of this fast-climbing aircraft from the airfield is represented by this blue arrow from the airfield to the aircraft. And the displacement of the slow-climbing aircraft from the airfield is given by the blue arrow from the airfield to this aircraft.

We’re asked how much greater the displacement of this fast-climbing aircraft is than the displacement of the slow-climbing aircraft. So, this is asking us to work out how much longer this blue arrow is than this one. To do this, we’re going to need to recall the two right-angled triangles that we identified in the diagram when answering the first part of the question. This time we need to find the difference between the length of this hypotenuse and the length of this one. Let’s label the length of the hypotenuse representing the displacement of the fast-climbing aircraft as 𝑑 one and that for the slow climbing aircraft as 𝑑 two. Then, what we’re trying to find in the second part of the question is 𝑑 one minus 𝑑 two.

The first step in doing this is to find the values of 𝑑 one and 𝑑 two individually. There are two possible approaches that we could use. One approach makes use of Pythagoras’s theorem, which says that for a right-angled triangle with a hypotenuse of length 𝑐 and other sides of lengths 𝑎 and 𝑏, that 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared.

In the triangles we’ve got in our diagram, we know that the length of one of the non-hypotenuse sides is 2500 meters for both of the two triangles. We also know that the vertical side of the blue triangle has a length of ℎ two and the vertical side of the pink triangle has a length of ℎ one. And in the first part of the question, we found the values of ℎ one and ℎ two. So, we could use our calculated values for ℎ one and ℎ two along with the other side length of 2500 meters to calculate the length of each triangle’s hypotenuse using this equation from Pythagoras’s theorem.

The second approach is to use a trigonometric equation, which says that for a right-angled triangle with an angle 𝜃, a side adjacent to that angle 𝜃 of length 𝑎, and a hypotenuse of length ℎ, that cos 𝜃 is equal to 𝑎 divided by ℎ. In the triangles in this diagram, we know the value of the angle 𝜃 in each case, and we know the length of the adjacent side 𝑎. So, we could rearrange this equation to make the hypotenuse ℎ the subject and use those values in it to calculate the hypotenuse of each of our triangles.

In this video, we’re going to use the second approach. While it’s perhaps a little more complicated than the first approach based on Pythagoras’s theorem, it has the advantage of not relying on any information that we calculated ourselves during the first part of the question. So even if we didn’t get the first part correct, it wouldn’t affect our result for this second part. We said that we were going to need to rearrange this equation to make ℎ the subject.

To do this, we first multiply both sides of the equation by ℎ. The ℎ’s on the right-hand side then cancel out. From here, we then divide both sides of the equation by cos 𝜃 so that the cos 𝜃 terms on the left-hand side cancel each other out. This leaves us with an equation that says the hypotenuse ℎ is equal to the adjacent side 𝑎 divided by cos 𝜃.

We can now apply this equation to each of the triangles from our diagram. Let’s start with the blue triangle that has an angle 𝜃 of 45 degrees, a side adjacent to this with a length of 2500 meters, so that’s our value for 𝑎, and its hypotenuse which is 𝑑 one is our value for ℎ. Substituting these values into this equation, we have that 𝑑 one is equal to 2500 meters divided by cos of 45 degrees. This works out as 3535.53 meters, where the ellipses show that this has further decimal places.

Now that we’ve got our value for 𝑑 one, the magnitude of the displacement of this fast-climbing aircraft, it’s time to do the same thing to find the value of 𝑑 two. So, in this case, 𝑑 two is our value for the hypotenuse ℎ. Our adjacent side 𝑎 is the same length of 2500 meters. And our angle 𝜃 is this value of 20 degrees. Using those values in this equation, we get that 𝑑 two is equal to 2500 meters divided by cos 20 degrees. This works out as 2660.44 meters. And like before, this result has further decimal places.

We’ve now worked out the values of both 𝑑 one and 𝑑 two. And we said that the answer to this part of the question is going to be 𝑑 one minus 𝑑 two, so let’s clear ourselves some space and work out this subtraction. Using the values that we have calculated, we have that 𝑑 one minus 𝑑 two is equal to 3535.53 meters minus 2660.44 meters. This works out as 875.09 meters. Finally, rounding this result to the nearest meter, like we’re asked, we get our answer of 875 meters.