### Video Transcript

A ball was thrown from the ground
at an angle π with a velocity π£. The ball reached a maximum height
of eight meters and a maximum range of 15 meters. Calculate the value of π. Let π equal 10 meters per second
squared. (A) 82.4 degrees, (B) 64.9 degrees,
(C) 46.8 degrees, and (D) 28.1 degrees.

In this question, weβve been told
that a ball is thrown from the ground at an angle π to the horizontal. We also know that the maximum
height of the ballβs trajectory is eight meters, and the horizontal distance the
ball travels between when itβs thrown and when it reaches the ground again is 15
meters. Based on this information, we need
to calculate the angle π between the horizontal and the direction of the ballβs
initial velocity π£.

To calculate π, letβs begin by
saying that the ball takes a time capital π to complete its trajectory. Since we will be ignoring any air
resistance acting on the ball, we can assume the trajectory is symmetric about the
midpoint and the ball is found at its maximum height, exactly halfway along its
journey, at a time of capital π over two.

Next, itβs worth noting that the
acceleration due to gravity is 10 meters per second squared. This will only act in the vertical
direction, since the gravitational force on the ball acts down towards the
ground. In other words, if the horizontal
direction is labeled π₯ and the vertical direction is labeled π¦, we can say that
the acceleration in the π¦-direction, π sub π¦, is equal to negative 10 meters per
second squared, or negative π as itβs been given to us in the question.

The reason this quantity is
negative is because we implicitly assumed that any motion in the upward direction is
positive when we labeled the maximum height of the ball as being positive eight
meters. The ball moved upward to get to a
maximum vertical displacement of eight meters, so a downward acceleration will have
a negative value. Weβve also been told in the
question that the ball is thrown with a velocity π£. We can find the horizontal and
vertical components of this velocity, so we can deal with the horizontal and
vertical motion of the ball separately.

Letβs recall that the ballβs
initial horizontal velocity, which we will call π£ sub π₯, is equal to π£ multiplied
by the cos of the angle π. Similarly, the initial vertical
velocity component, π£ sub π¦, is equal to π£ multiplied by the sin of π. The reason we do this is because we
can now study the horizontal and vertical motion of the ball individually, which
will make our calculations simpler to deal with.

For example, letβs recall that for
an object moving at a constant velocity, the velocity of that object is equal to the
displacement of the object divided by the time taken to travel this
displacement. We can apply this equation to the
horizontal motion of the ball because in this direction, the velocity of the ball is
indeed constant.

Hence, clearing some space to work,
we find that the horizontal velocity of our ball, π£ times the cos of π, is equal
to the horizontal displacement π sub π₯ divided by the time taken to move this
distance, which weβve called capital π. Now we know the value of π sub π₯
β itβs 15 meters β but we wonβt substitute this in just yet.

Next, letβs apply some kinematic
equations to the horizontal and vertical motion of the ball. Letβs recall that the kinematic
equations can only be applied to objects moving with a constant acceleration. Our ball fits this criterion, as it
has a constant downward acceleration due to gravity. The first kinematic equation we can
recall is this one. The final velocity of an object, π£
sub f, is equal to the initial velocity, π£ sub i, plus the acceleration, π,
multiplied by the time for which it moves π‘.

We can apply this to the vertical
motion of the ball between when it is thrown and when it reaches its maximum
height. The final vertical velocity of the
ball in this scenario is actually zero. When the ball is at its maximum
height, it must have exactly zero vertical velocity. Because if this was not true, then
the ball would either continue moving upward or not have got this high, and the
maximum height would be something else. So, π£ sub f π¦ is equal to the
initial vertical velocity, which is π£ times the sin of π, plus the acceleration,
which is negative π, multiplied by the time for which the ball moves.

Since weβre only studying the
motion up until the ball reaches the maximum height, this time is going to be
capital π divided by two, half the time for the entire motion. We can now rearrange these two
equations weβve found so we can solve for π£ in both cases. For this equation, where recall π£
sub f π¦ is zero, we find π£ is equal to π times capital π divided by two times
the sin of π. And the first equation can be
rearranged so it reads π£ is equal to π sub π₯ divided by capital π times the cos
of π.

Now, we can equate the two
right-hand sides of these equations as they are both equal to π£. When we do this, we can rearrange
once again. So, our equation reads π sub π₯
multiplied by the sin of π divided by the cos of π is equal to π times capital π
squared all divided by two. The reason we rearrange like this
is because we can recall that sin π divided by cos π is equal to tan π.

Next, we can apply another
kinematic equation to the vertical motion of the ball. This equation tells us that the
displacement of an object is equal to the initial velocity multiplied by the time of
travel plus one-half times the acceleration times the time squared.

For the ball falling from its
maximum height back to ground, the vertical displacement is negative π sub π¦, the
initial vertical velocity is zero, the acceleration is negative π, and the time is
capital π divided by two. Canceling out the term which is
zero, due to the initial velocity being zero, and canceling the negative signs on
both remaining terms, we find that π sub π¦ is equal to one-half times π times
capital π squared over four.

Another way to write the right-hand
side of this equation is as one-fourth times the quantity one-half multiplied by π
times capital π squared. The reason we do this is because
our equation from earlier also has the quantity one-half π times capital π squared
as its right-hand side. This means we can equate the two
left-hand sides if we remember to account for this factor of one-fourth. We find that π sub π¦ is equal to
one-fourth multiplied by π sub π₯ times the tan of π.

At this point, we know all the
quantities in this equation apart from π, so we can rearrange to solve for it. We find that tan of π is equal to
four times π sub π¦ divided by π sub π₯. So, π is equal to the inverse tan
of this quantity. We finally substitute in π sub π¦
is equal to eight meters and π sub π₯ is equal to 15 meters. The units of meters in the
numerator and denominator cancel, which is great because we need a unitless number
in order to be able to find the inverse tangent of it. When we evaluate this expression,
we find π to be equal to 64.88 and so on degrees. To one decimal place, this becomes
64.9 degrees.

Looking back at our answer options,
we see this corresponds to option (B). Therefore, weβve found that the
angle between the ground and the initial velocity of the ball is 64.9 degrees.