# Video: Finding the Components of the Projection Velocity of a Particle and the Maximum Height

A particle projected with a velocity (𝑢𝐢 + 𝑢𝐣) m/s from a fixed point 𝑂 in a horizontal plane landed at a point in the same plane 360 m away. Find the value of 𝑢 and the projectile’s path’s greatest height ℎ. Take 𝑔 = 9.8 m/s².

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### Video Transcript

A particle projected with a velocity 𝑢𝐢 plus 𝑢𝐣 metres per second from a fixed point 𝑂 in a horizontal plane landed at a point in the same plane 360 metres away. Find the value of 𝑢 and the projectile’s path’s greatest height ℎ. Take 𝑔 equals 9.8 metres per square second.

When dealing with projectile motion, it can be really useful to sketch a diagram. We know that the motion of the particle will look a little bit like this inverted parabola. We’re also told that the horizontal and vertical components of the initial velocity are both equal to 𝑢. So, we can drop in a right-angle triangle as shown. We also know that the particle lands 360 metres away. So, that means the final horizontal displacement of the particle is 360 metres. We also have acceleration due to gravity. That acts downwards.

So, in order to answer these questions, we’re going to use the equations of constant acceleration, sometimes called the SUVAT equations. But we’re going to separate these into the horizontal and vertical motion of the particle. We’ll begin with the horizontal motion. Initial velocity in the horizontal direction, we’re told, is equal to 𝑢. We also know that horizontally there is no acceleration. So, that’s zero. And this means that its horizontal velocity, until it reaches the ground at least, will always be equal to 𝑢. We’re going to use the final horizontal displacement to help us calculate a few other bits of information. And we know that’s equal to 360. We don’t yet know anything about the time.

We’ll now consider the vertical motion. The initial velocity vertically is 𝑢. We’re told that acceleration due to gravity, which we’ve labelled 𝑔, is 9.8. Now, that acts in the opposite direction that we defined the vertical velocity to be in. So, we’re going to say that’s negative 9.8. We don’t yet know anything about the vertical velocity at a given time. But we do know that when the particle hits the ground, its vertical displacement must be equal to zero. And we don’t yet know the time at which this occurs.

We do, however, have enough information to help us find the value of 𝑢. Let’s list those SUVAT equations. We have 𝑣 equals 𝑢 plus 𝑎𝑡, 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared, 𝑠 equals a half 𝑢 plus 𝑣 times 𝑡, 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠, and 𝑠 equals 𝑣𝑡 minus a half 𝑎𝑡 squared. We’re going to begin by looking at the horizontal motion. And we’re going to use the third equation, 𝑠 equals a half 𝑢 plus 𝑣 times 𝑡. This will tell us the value of 𝑡, the time that the particle is in motion.

We get 360 equals a half times 𝑢 plus 𝑢 times 𝑡, or simply 360 equals 𝑢 times 𝑡. By dividing through by 𝑢, we find that 𝑡 is equal to 360 over 𝑢. This remains consistent in the vertical direction. Remember, we’re looking at the point that the particle hits the ground again. This time we’re going to use the second equation. We’re not really interested in the vertical velocity. We get zero equals 𝑢𝑡 plus a half times negative 9.8 𝑡 squared.

A half times negative 9.8 is negative 4.9. And then, we can replace 𝑡 with 360 over 𝑢. These two 𝑢s cancel. We then add 4.9 times 360 over 𝑢 squared to both sides then multiply through by 𝑢 squared and divide through by 360. So, we get 𝑢 squared is equal to 4.9 times 360, which means 𝑢 is equal to the square root of 4.9 times 360, which is equal to 42. Now, here, we were only interested in the positive square root since we define the initial horizontal velocity to be positive. And so, 𝑢 must be equal to 42 metres per second.

The second part of this question asks us to find the projectile’s path’s greatest height. Now, there are a couple of ways that we can do this, but we can redefine a few elements in the vertical direction. We’re no longer interested in the horizontal motion. We know that 𝑢 is equal to 42. The height we’re interested in is ℎ. We don’t know the time at which this occurs, although we could work it out by finding the total time taken and then halving it.

Alternatively, we recognise that at the moment that the particle reaches its highest height, its velocity in the vertical direction is zero. And so, we find the equation which does not include 𝑡. It’s 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠. Substituting what we know about our particle, we get zero squared equals 42 squared plus two times negative 9.8 times ℎ. 42 squared is 1764. And two times 9.8 is 19.6. So, we add 19.6ℎ to both sides. And all that’s left to do is divide through by 19.6. That gives us a value of 90. And so, we can say that 𝑢 is 42 metres per second. And the projectile’s path’s greatest height ℎ is 90, or 90 metres.