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

If a particle was projected from the origin π on a horizontal ground with initial velocity (17π’ + 11π£) m/s, write its position vector at time π‘ 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 π on a horizontal ground with initial velocity 17π’ plus 11π£ metres per second, write its position vector at time π‘ seconds. Consider the acceleration due to gravity to be 9.8 metres per square second.

Weβ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 letβs call that π  β at time π‘ seconds. And so weβre going to use one of the equations of constant acceleration. Weβre going to use the equation π  equals π’π‘ plus a half ππ‘ squared. Now, we can convert this into vector form. We can say that π , π’, and π are vectors.

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

Letβs distribute each of these set of parentheses. And we get π  equals 17π‘π’ plus 11π‘π£ minus 4.9π£π‘ squared. And we now gather the vertical components. And that gives us π  equals 17π‘π’ plus 11π‘ minus 4.9π‘ squared π£. This is measured in metres. And so weβre done. We found a position vector at time π‘ seconds. Its π  equals 17π‘π’ plus 11π‘ minus 4.9π‘ squared π£ metres.