Video Transcript
A particle projected from the
origin 𝑂 passed horizontally through a point with a position vector of 10𝐢 plus
10𝐣 meters, where 𝐢 and 𝐣 are horizontal and vertical unit vectors,
respectively. Determine the velocity with which
the particle left 𝑂, 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 𝑂. We don’t know its initial velocity,
so let’s call that 𝑢. We can split that initial velocity
into its horizontal and vertical components. Let’s call the horizontal component
for the initial velocity 𝑢 sub 𝑥 and the vertical component 𝑢 sub 𝑦. 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 10𝐢 plus 10𝐣 meters. And then we can see the
acceleration due to gravity is 9.8 meters per second in the downward direction.
What we’re 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 𝑢 sub 𝑥. This is essentially what we’re
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 𝑢 sub 𝑥. We’re interested in a horizontal
displacement of 10 meters. And we don’t know the time at which
this occurs. We defined the initial vertical
velocity to be 𝑢 𝑦. And then the vertical component for
acceleration acts in the opposite direction. It’s acceleration due to
gravity. And we define that to be negative
9.8.
We’re interested in the highest
point of our ark. And we said this occurs when the
vertical velocity is equal to zero. And that’s 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 don’t know the
time at which this occurs. We have actually got enough
information to workout the vertical component for the initial velocity. We’re looking to find 𝑢. We know 𝑎, 𝑣, and 𝑠. And so we use one of our equations
of constant acceleration. This is sometimes called SUVAT
equations because of the acronym the letters make.
We’re not really interested in the
time at this point, so we’re going to use the equation 𝑣 squared equals 𝑢 squared
plus two 𝑎𝑠. 𝑣 is equal to zero. 𝑢 is equal to 𝑢 𝑦. Remember, that’s what we’re trying
to find, so we write 𝑢 𝑦 squared. And then we want two times 𝑎 times
𝑠. That’s two times negative 9.8 times
10. This simplifies to zero equals 𝑢
𝑦 squared minus 196. We add 196 to both sides. And then we’re going to square root
both sides of the equation. The square root of 196 is 14. And we’re 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 𝑢 𝑦. But how do we find 𝑢 𝑥? Well, we are able now to calculate
the time at which the particle reaches this position. We’re going to continue looking at
the vertical motion, this time using the formula 𝑣 equals 𝑢 plus 𝑎𝑡. Substituting what we know about our
vertical motion, and we get zero equals 14 minus 9.8𝑡. We add 9.8𝑡 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 𝑢 𝑥 in the horizontal direction.
This time we’re going to use this
equation, 𝑠 equals a half 𝑢 plus 𝑣𝑡. We substitute what we know about
the horizontal motion. And we get 10 equals a half times
𝑢 sub 𝑥 plus 𝑢 sub 𝑥 times ten-sevenths. This simplifies on the right-hand
side to 𝑢 sub 𝑥 times ten-sevenths, and we divide through by ten-sevenths. And we find 𝑢 sub 𝑥 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 𝑂 is seven 𝐢 plus 14𝐣 meters per
second.
