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

Video Transcript

If a particle was projected from the origin 𝑂 on a horizontal ground with initial velocity 17𝐢 plus 11𝐣 metres per second, write its position vector at time 𝑡 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 — let’s call that 𝑠 — at time 𝑡 seconds. And so we’re going to use one of the equations of constant acceleration. We’re going to use the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. Now, we can convert this into vector form. We can say that 𝑠, 𝑢, and 𝑎 are vectors.

Lets substitute what we know about our particle into this formula. 𝑠 is equal to 𝑢 times 𝑡. And we know that 𝑢 is 17𝐢 plus 11𝐣. And then we have a half 𝑎𝑡 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 17𝐢 plus 11𝐣𝑡 plus a half times zero 𝐢 minus 9.8𝐣𝑡 squared.

Let’s distribute each of these set of parentheses. And we get 𝑠 equals 17𝑡𝐢 plus 11𝑡𝐣 minus 4.9𝐣𝑡 squared. And we now gather the vertical components. And that gives us 𝑠 equals 17𝑡𝐢 plus 11𝑡 minus 4.9𝑡 squared 𝐣. This is measured in metres. And so we’re done. We found a position vector at time 𝑡 seconds. Its 𝑠 equals 17𝑡𝐢 plus 11𝑡 minus 4.9𝑡 squared 𝐣 metres.