### Video Transcript

A ball is launched diagonally upward and then returns to the ground. The change of the vertical displacement of the ball with time is shown in the graph. The ball lands a 2.4-meter horizontal distance from where it was launched and reaches a maximum vertical displacement of 4.9 meters. The ball has a constant horizontal speed throughout its flight. What is the initial vertically upward speed of the ball? What is the horizontal speed of the ball? At what angle above the horizontal is the ball launched? Give your answer to the nearest degree.

Letβs start with the first part of this question. In this question, we have a ball thatβs launched diagonally upward with an initial speed that we will call π, and this is an angle above the horizontal that we will call π. Throughout its motion, there will be a downward force acting on the ball. This is caused by gravity and it has a magnitude of the mass of the ball, which we will call π, multiplied by the acceleration due to gravity, which is π.

Importantly, we are told that the ball has a constant horizontal speed throughout its flight. This means that thereβs no force acting horizontally on the ball, and therefore the only force acting on the ball is gravity. The ball will have a curved path through the air until it lands, which we are told is some horizontal distance from where it was launched. And for now, weβll call this horizontal distance capital π
.

Letβs take a look at the initial velocity of the ball. Now, this is a vector quantity, meaning it has a magnitude, which is the initial speed of the ball, and a direction, which is given to us by the launch angle of the ball above the horizontal. We can see that it has a horizontal component that we will call π sub π₯ and a vertical component that we will call π sub π¦. And as weβve already seen, π sub π₯ is constant. So, if we were to look at the ball at any point through its trajectory, say at any one of these points, we would see that the horizontal speed of the ball does not change. However, if we were to mark the vertical component of the ballβs velocity, we would see that it decreases over time.

Starting with its initial positive or upwards value and then decreasing over time until halfway through the motion, the ball instantaneously has zero vertical velocity. After this, the ballβs vertical velocity becomes negative and the ball will begin moving downwards until it lands. This point when the ball has zero vertical velocity is also when itβs at its maximum altitude, and we will call this vertical displacement β. The question tells us two things. First, it tells us that the ball lands 2.4 meters horizontally distant from where it was launched. So, we can write that π
is equal to 2.4 meters. Second, it tells us that the ball reaches a maximum vertical displacement of 4.9 meters, so β is equal to 4.9 meters.

We are also given a graph, which shows time on the horizontal axis and vertical displacement on the vertical axis. From the graph, we can again see that the maximum vertical displacement of the ball is around 4.9 meters, and this happens at exactly one second through its motion. In fact, we can see that the ball returns to a vertical displacement of zero meters at exactly 2.0 seconds. This is the point that the ball lands, so we can say that it has a time of flight of two seconds. We will denote this with a capital π, and weβll keep a note of it over here on the right.

Okay, so onto the first part of the question, which asks, what is the initial vertically upward speed of the ball? Letβs clear some space so we can work this out. We can actually estimate the initial vertically upward speed of the ball by looking at the first two points on our graph, these two over here.

Recall that the speed of an object is equal to the change in its position divided by the change in time. In our case, the initial vertically upward speed of the ball π π¦ is going to approximately equal the change in position between those two points on the graph divided by the change in time between those two points. The second of these points has a vertical displacement of approximately 0.95 meters, and the first point is at exactly zero meters. The second point is recorded at a time of exactly 0.1 seconds, and the first point is at the very beginning of the motion, so itβs at a time of zero seconds.

Evaluating this, we get an approximate answer for the initial vertically upward speed of the ball of 9.5 meters per second. This seems reasonable, but there is a way to do this without using the graph at all. And thatβs by using the following equation for projectile motion, which tells us that the maximum vertical displacement of a projectile is equal to its initial speed multiplied by the sine of its launch angle above the horizontal all squared divided by two times the acceleration due to gravity. But how can we relate this to the initial vertically upward speed of the ball?

Well, if we look again at our diagram of the initial speed of the ball, we can relate the initial speed of the ball to its initial vertical speed using trigonometry, where we see that the initial vertically upward speed of the ball is equal to the ballβs initial speed multiplied by the sine of its launch angle above the horizontal, or π π¦ is equal to π sin π. And we see that this actually appears in our equation for the maximum vertical displacement of the ball. So, we can rewrite this as β is equal to π π¦ squared divided by two π.

We now have to rearrange this to make π π¦ the subject of the equation if we want to find the initial vertically upward speed of the ball. First, weβll multiply both sides by two π, where we see that these two πβs on the right cancel. Next, we take the square root of both sides, and the square root of a squared number is just the number itself, so the square root of π π¦ squared is just π π¦. This gives us our expression for the initial vertically upward speed of the ball. Letβs write this a little bit neater. π π¦ is equal to the square root of two π β.

Now, we can substitute our known value of β and recall that π is equal to 9.8 meters per second squared. These are both in SI units, so we donβt need to convert either of them before continuing. Substituting these in, we get π π¦ is equal to the square root of two multiplied by 9.8 meters per second squared multiplied by 4.9 meters.

Now, for the units, we have meters per second squared multiplied by meters, which gives us meters squared divided by seconds squared. When we take the square root of this, we get meters per second, which we would expect for speed. Using our calculator to evaluate this expression, we get π π¦ is equal to 9.8 meters per second, and this is very close to the value we calculated by looking at our graph. So, the initial vertically upward speed of the ball is 9.8 meters per second.

Now, letβs take a look at the second part of this question. The second part of the question asks us, what is the horizontal speed of the ball?

Now, this is relatively straightforward given that we know exactly how far the ball has traveled in the horizontal direction and the amount of time itβs taken to do so. Because the ball has a constant horizontal speed, this speed is equal to the change in horizontal position of the ball divided by the change in time. Or, in our case, π π₯ is equal to π
divided by π. Substituting our known values for π
and π into this, we get π π₯ is equal to 2.4 meters divided by 2.0 seconds. And we should note that units of meters in the numerator and seconds in the denominator will give us overall units of meters per second, which is what we expect for a speed. Evaluating this expression, we get π π₯ is equal to 1.2 meters per second. So, the horizontal speed of the ball is 1.2 meters per second.

Finally, letβs move on to the last part of this question. The final part of this question asks, at what angle above the horizontal is the ball launched? Give your answer to the nearest degree.

We can actually work this out by looking at our diagram of the initial speed of the ball, where because these form a right angle triangle, we can use trigonometry to find the angle above the horizontal that the ball was launched, where we see that tan π is equal to π π¦ divided by π π₯. Taking the inverse tan of both sides and noting that the inverse tan of the tan of π is just π, we get our expression for the launch angle of the ball. All thatβs left to do is to substitute our known values of π π¦ and π π₯ into this equation, which gives us π is equal to the inverse tan of 9.8 meters per second divided by 1.2 meters per second. Evaluating this, π is equal to 83.0189 and so on degrees. The question asked us to give our answer to the nearest degree, so π to the nearest degree is just 83 degrees.

The angle that the ball is launched above the horizontal is 83 degrees to the nearest degree.