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Question Video: Calculating the Launch Speed of a Projectile from the Motion of another Projectile Physics • 9th Grade

Two projectiles are launched along the same horizontal direction from the same point at the same time. Projectile 𝑃₁ is launched at an angle of 65Β° above the horizontal at a speed of 2.5 m/s. Projectile 𝑃₂ is launched at an angle of 15Β° above the horizontal. The projectiles land at the same point. At what speed was projectile 𝑃₂ launched?

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Video Transcript

Two projectiles are launched along the same horizontal direction from the same point at the same time. Projectile 𝑃 one is launched at an angle of 65 degrees above the horizontal at a speed of 2.5 meters per second. Projectile 𝑃 two is launched at an angle of 15 degrees above the horizontal. The projectiles land at the same point. At what speed was projectile 𝑃 two launched?

Let’s start with the diagram of projectile 𝑃 one. 𝑃 one is launched at an angle that we will call πœƒ one with an initial speed that we will call 𝑉 one. The only force acting on 𝑃 one is from the object’s weight which has a magnitude of the mass of projectile one, which we will call π‘š one, multiplied by the acceleration due to gravity, which is 𝑔. This force from gravity acts downwards, so the projectile’s path through the air is curved. The projectile then lands at a position that has a horizontal distance from the launch point of 𝑅 one.

Now, let’s look at projectile 𝑃 two. 𝑃 two is launched with an initial speed that we will call 𝑉 two at an angle above the horizontal that we will call πœƒ two. Again, the only force acting on the projectile is the object’s weight, which has a magnitude of the mass of projectile 𝑃 two, which we will call π‘š two, multiplied by the acceleration due to gravity, which is 𝑔. Just like projectile 𝑃 one, 𝑃 two will have a curved path through the air. And we are told in the question that the projectiles land at the same point. The horizontal distance from the launch position of 𝑃 two to the position where it lands we will call 𝑅 two and is equal to 𝑅 one.

The question asks us to find the speed that projectile 𝑃 two was launched at. So we need to find 𝑉 two. We will start with the equation for the horizontal range of the projectile, which states that the horizontal range of a projectile 𝑅 is equal to two multiplied by its initial speed squared, which is 𝑉 squared, multiplied by cos πœƒ multiplied by sin πœƒ, where πœƒ is the projectile’s launch angle above the horizontal, divided by 𝑔, the acceleration due to gravity. So for the first projectile, we can write 𝑅 one is equal to two multiplied by 𝑉 one squared multiplied by the cos of πœƒ one multiplied by the sin of πœƒ one divided by 𝑔. And for the second projectile, 𝑃 two, we can write 𝑅 two is equal to two multiplied by 𝑉 two squared multiplied by the cos of πœƒ two multiplied by the sin of πœƒ two divided by 𝑔.

Now, what’s interesting here is that neither of these equations depend on the mass of the projectiles. And that’s because they’re both experiencing the same acceleration, which is the acceleration due to gravity. We know that the projectiles land at the same point, which means that 𝑅 one is equal to 𝑅 two. So these expressions on the right are equal to each other.

The question asks us to find the speed that the projectile 𝑃 two was launched at. In other words, we need to find an expression for 𝑉 two. We will do this by rearranging the equation on the right. We will start by canceling common terms in this equation. First, we will divide both sides of the equation by two. This will cancel the two on either side of the equation. Next, we’ll multiply both sides of the equation by 𝑔. This will cancel the one over 𝑔 factor on either side. Next, we will divide both sides by the cos of πœƒ two and also by the sin of πœƒ two. And we notice that these cos and sin terms on the right cancel.

Finally, we take the square root of both sides, and the square root of 𝑉 two squared is just 𝑉 two. Similarly, we can bring 𝑉 one outside of the square root on the left. This is because the square root of 𝑉 one squared multiplied by the other terms on the left is equal to the square root of 𝑉 one squared multiplied by the square root of the other terms. And as we know, the square root of 𝑉 one squared is just 𝑉 one. So this works out as 𝑉 one multiplied by the square root of the other terms, which gives our final expression for 𝑉 two.

Let’s write this out a little bit more neatly. 𝑉 two is equal to 𝑉 one multiplied by the square root of the cos of πœƒ one multiplied by the sin of πœƒ one divided by the cos of πœƒ two multiplied by the sin of πœƒ two. All that’s left for us to do is to substitute our values of 𝑉 one, πœƒ one, and πœƒ two into this equation. The question tells us that projectile 𝑃 one is launched at an angle of 65 degrees above the horizontal. So πœƒ one is equal to 65 degrees. It also tells us that projectile 𝑃 one is launched at a speed of 2.5 meters per second. So 𝑉 one is equal to 2.5 meters per second.

The question goes on to say that projectile 𝑃 two is launched at an angle of 15 degrees above the horizontal. So πœƒ two is equal to 15 degrees. Substituting these into our equation, we get 𝑉 two is equal to 2.5 meters per second multiplied by the square root of the cos of 65 degrees multiplied by the sin of 65 degrees divided by the cos of 15 degrees multiplied by the sin of 15 degrees. Evaluating this gives us an answer of 𝑉 two is equal to 3.1 meters per second.

Importantly, because the cosine and sine of an angle have no units, the units of 𝑉 two will be whatever the units of 𝑉 one were. In this case, 𝑉 one was given to us in SI units of meters per second. So our answer is also in SI units of meters per second. So there’s no need for us to convert our answer into any other units. So the speed that projectile 𝑃 two was launched at is 3.1 meters per second.

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