# Video: Solving Problems on Projectile Motion

Find, to one decimal place, a projectile’s speed 3 seconds after it was thrown, given that it flew a total horizontal distance of 282 m for 5 seconds before hitting the ground. Take 𝑔 = 9.8 m/s².

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### Video Transcript

Find, to one decimal place, a projectile’s speed three seconds after it was thrown, given that it flew a total horizontal distance of 282 metres for five seconds before hitting the ground. Take 𝑔 to be equal to 9.8 metres per square second.

Let’s begin by drawing a sketch of what’s going on here. The motion of the particle looks a little something like this: it’s an inverted parabola. It covers a total distance of 282 metres horizontally. Acceleration due to gravity acts in the vertical direction; that’s downwards. And then, we have the initial speed. Now, we don’t know the initial speed. So, what we’re going to do is split it into the horizontal and vertical components of the initial speed. So horizontally, its initial speed is 𝑢 sub 𝑥. And vertically, its initial speed is 𝑢 sub 𝑦 upwards.

We’re now going to consider the motion of particle in both the horizontal and vertical direction. Our aim is to find the initial speed in both directions. Let’s begin horizontally. Its initial speed is 𝑢 sub 𝑥. We know that the acceleration horizontally is zero. And this means its horizontal velocity remains unchanged. So its velocity at any point is still 𝑢 sub 𝑥. It travels a total distance of 282 metres and it takes five seconds to do so. We’re going to use one of the equations of constant acceleration to find the initial velocity of the particle.

We’re going to use 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. And the reason behind using this one in particular will become evident in a moment. Horizontally, the total displacement is 282. 𝑢𝑡 is 𝑢 sub 𝑥 times five. And then, a half 𝑎𝑡 squared is a half times zero times five squared. But of course, that’s simply zero. So, we have 282 equals 𝑢 sub 𝑥 times five. And we can solve this equation by dividing through by five. And we get 𝑢 sub 𝑥 to be equal to 282 over five or 282 over five metres per second.

Let’s repeat this process for the vertical motion. We defined the initial velocity in the vertical direction to be 𝑢 𝑦. Acceleration due to gravity acts in the opposite direction as negative 9.8. We don’t know the velocity at a given point, although we could potentially work it out. But we do know that the vertical displacement is zero. It ends up at exactly the same height as it started. And once again, we know that this takes five seconds. We’ll use the same formula here. This time, 𝑠 is zero. 𝑢𝑡 is 𝑢 sub 𝑦 times five. Then, a half 𝑎𝑡 squared is a half times negative 9.8 times five squared. And this equation simplifies to zero equals five 𝑢 sub 𝑦 minus 245 over two. We’re going to add 245 over two to both sides to get 245 over two equals five 𝑢 sub 𝑦. And then, we divide through by five. And we find 𝑢 sub 𝑦 to be equal to 49 over two.

Now that we have all this information, we can work out the speed three seconds after the projectile was thrown. Now, remember, speed is the magnitude of the velocity. So, we’ll work out the horizontal and vertical velocity and then find the magnitude. This time, we use the equation 𝑣 equals 𝑢 plus 𝑎𝑡. Now, in fact, we don’t need to do this in the horizontal direction since we said acceleration is zero. So, 𝑣 𝑥, the velocity at any point, is 282 over five.

We’re looking to find the speed three seconds after it was thrown. Now, we don’t know the displacement. But we do know that 𝑡 is equal to three. So, 𝑣 sub 𝑦 is 𝑢 plus 𝑎𝑡. That’s 49 over two minus 9.8 times three, which is negative 49 tenths. The speed is the square root of the sum of the squares of each of the horizontal and vertical components for velocity. So, it’s the square root of 282 over five squared plus negative 49 over 10 squared. That’s 56.612 and so on, which correct to one decimal place is 56.6 metres per second.

And so, the speed three seconds after the projectile was thrown is 56.6 metres per second.