### 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.

Weβre 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 letβs call that π β at time π‘ seconds. And so weβre going to use one of
the equations of constant acceleration. Weβre going to use the equation π
equals π’π‘ plus a half ππ‘ 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 ππ‘
squared. Now, acceleration in the horizontal
direction is zero. But in the vertical direction, it
acts against the direction of the original velocity. So itβs 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.

Letβs 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 weβre done. We found a position vector at time
π‘ seconds. Its π equals 17π‘π’ plus 11π‘
minus 4.9π‘ squared π£ metres.