### Video Transcript

A particle projected from the
origin π passed horizontally through a point with a position vector of 10π’ plus
10π£ meters, where π’ and π£ are horizontal and vertical unit vectors,
respectively. Determine the velocity with which
the particle left π, considering the acceleration due to gravity to be 9.8 meters
per square second.

It can be really useful to simply
begin by sketching a diagram of the motion of the particle. We know that the particle is
projected from the origin π. We donβt know its initial velocity,
so letβs call that π’. We can split that initial velocity
into its horizontal and vertical components. Letβs call the horizontal component
for the initial velocity π’ sub π₯ and the vertical component π’ sub π¦. There is a moment when the particle
is passing horizontally through a point. This must be at the highest point
the particle reaches. We know this occurs when the
vertical velocity is equal to zero. And we know, at this point, the
position vector is 10π’ plus 10π£ meters. And then we can see the
acceleration due to gravity is 9.8 meters per second in the downward direction.

What weβre going to do is split the
motion up into its horizontal components and its vertical components. Horizontally, we defined the
initial velocity to be equal to π’ sub π₯. This is essentially what weβre
going to be finding. There is no acceleration that acts
on this particle in the horizontal direction. And this means that its horizontal
velocity never changes. So its horizontal velocity at any
point will also be π’ sub π₯. Weβre interested in a horizontal
displacement of 10 meters. And we donβt know the time at which
this occurs. We defined the initial vertical
velocity to be π’ π¦. And then the vertical component for
acceleration acts in the opposite direction. Itβs acceleration due to
gravity. And we define that to be negative
9.8.

Weβre interested in the highest
point of our ark. And we said this occurs when the
vertical velocity is equal to zero. And thatβs because, at this point,
the particle changes direction. So there is an instantaneous moment
and its velocity is equal to zero. The vertical component for the
displacement at this point is 10. And yet again, we donβt know the
time at which this occurs. We have actually got enough
information to workout the vertical component for the initial velocity. Weβre looking to find π’. We know π, π£, and π . And so we use one of our equations
of constant acceleration. This is sometimes called SUVAT
equations because of the acronym the letters make.

Weβre not really interested in the
time at this point, so weβre going to use the equation π£ squared equals π’ squared
plus two ππ . π£ is equal to zero. π’ is equal to π’ π¦. Remember, thatβs what weβre trying
to find, so we write π’ π¦ squared. And then we want two times π times
π . Thatβs two times negative 9.8 times
10. This simplifies to zero equals π’
π¦ squared minus 196. We add 196 to both sides. And then weβre going to square root
both sides of the equation. The square root of 196 is 14. And weβre not interested in the
negative square root because we define the initial velocity to be in the positive
direction.

So we found the value of π’ π¦. But how do we find π’ π₯? Well, we are able now to calculate
the time at which the particle reaches this position. Weβre going to continue looking at
the vertical motion, this time using the formula π£ equals π’ plus ππ‘. Substituting what we know about our
vertical motion, and we get zero equals 14 minus 9.8π‘. We add 9.8π‘ to both sides and then
divide through by 9.8. And that gives us ten-sevenths. So we now have enough information
to work out π’ π₯ in the horizontal direction.

This time weβre going to use this
equation, π equals a half π’ plus π£π‘. We substitute what we know about
the horizontal motion. And we get 10 equals a half times
π’ sub π₯ plus π’ sub π₯ times ten-sevenths. This simplifies on the right-hand
side to π’ sub π₯ times ten-sevenths, and we divide through by ten-sevenths. And we find π’ sub π₯ to be equal
to seven. We now know the horizontal and
vertical components for the initial velocity. And so we can say that the initial
velocity with which the particle left π is seven π’ plus 14π£ meters per
second.