Video: Finding the Position Vector of a Projectile given the Initial Velocity of the Projectile as a Vector

If a particle was projected from the origin 𝑂 on a horizontal ground with initial velocity (17𝐒 + 11𝐣) m/s, write its position vector at time 𝑑 seconds. Consider the acceleration due to gravity to be 9.8 m/sΒ².

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Video Transcript

If a particle was projected from the origin 𝑂 on a horizontal ground with initial velocity 17𝐒 plus 11𝐣 metres per second, write its position vector at time 𝑑 seconds. Consider the acceleration due to gravity to be 9.8 metres per square second.

We’re given the initial velocity of the particle in a vector form. Its horizontal component is 17 metres per second and its vertical component is 11 metres per second. We want to find its position vector β€” let’s call that 𝑠 β€” at time 𝑑 seconds. And so we’re going to use one of the equations of constant acceleration. We’re going to use the equation 𝑠 equals 𝑒𝑑 plus a half π‘Žπ‘‘ squared. Now, we can convert this into vector form. We can say that 𝑠, 𝑒, and π‘Ž are vectors.

Lets substitute what we know about our particle into this formula. 𝑠 is equal to 𝑒 times 𝑑. And we know that 𝑒 is 17𝐒 plus 11𝐣. And then we have a half π‘Žπ‘‘ squared. Now, acceleration in the horizontal direction is zero. But in the vertical direction, it acts against the direction of the original velocity. So it’s negative 9.8. And so we find that 𝑠 is equal to 17𝐒 plus 11𝐣𝑑 plus a half times zero 𝐒 minus 9.8𝐣𝑑 squared.

Let’s distribute each of these set of parentheses. And we get 𝑠 equals 17𝑑𝐒 plus 11𝑑𝐣 minus 4.9𝐣𝑑 squared. And we now gather the vertical components. And that gives us 𝑠 equals 17𝑑𝐒 plus 11𝑑 minus 4.9𝑑 squared 𝐣. This is measured in metres. And so we’re done. We found a position vector at time 𝑑 seconds. Its 𝑠 equals 17𝑑𝐒 plus 11𝑑 minus 4.9𝑑 squared 𝐣 metres.

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