Question Video: Analyzing Projectile Vertical and Horizontal Speed | Nagwa Question Video: Analyzing Projectile Vertical and Horizontal Speed | Nagwa

Question Video: Analyzing Projectile Vertical and Horizontal Speed Physics • First Year of Secondary School

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An object is set in motion by an initial force 𝐹 that acts diagonally upward, as shown in the diagram. The object undergoes projectile motion. Which of the graphs (a), (b), (c), and (d) shows the changes in the vertical speed of the object between leaving the ground and returning to the ground? Which of the graphs (e), (f), (g), and (h) shows the changes in the horizontal speed of the object between leaving the ground and returning to the ground?

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Video Transcript

An object is set in motion by an initial force 𝐹 that acts diagonally upward, as shown in the diagram. The object undergoes projectile motion. Which of the graphs (a), (b), (c), and (d) shows the changes in the vertical speed of the object between leaving the ground and returning to the ground?

In our diagram, we see our projectile, represented by this gray dot, moving through its trajectory. We’re told that the force 𝐹 that originally acts on this object to get it in motion acts diagonally upward. We know then that our object will have both a vertical as well as a horizontal speed. In this first part of our question, we’re concentrated on that vertical aspect of our object’s motion. We want to know which of these four graphs correctly shows the object’s vertical speed over time.

Let’s first of all notice that the first two graphs, graphs (a) and (b), have the vertical speed of this object being zero at a time of zero. If we think though about our actual projectile being acted on by this force, we realize that its vertical speed at this moment will actually have its maximum value. At this initial instant, the object is moving upward relatively quickly. But then, as the object ascends, its vertical speed gets smaller and smaller. Finally, at this moment here, the object’s vertical speed is zero. For just an instant, it’s moving neither up nor down. Then, of course, the object begins to fall back to earth, and its vertical speed increases.

Note that we are here talking about speed rather than velocity. We’re considering a scalar quantity whose values are always nonnegative. Getting back to answer options (a) and (b) though, we see that neither of these options can be correct because they both indicate the object’s vertical speed of zero at a time of zero. We’ve seen that what actually happens is that vertical speed is maximum at this moment.

Now we have to decide between graphs (c) and (d). To help us do that, let’s note that this object is indeed undergoing projectile motion. This means that a set of equations for describing projectile motion apply to our object. Let’s clear some space on screen and consider this kinematic equation of projectile motion. This equation is true whenever the acceleration an object undergoes is constant. That’s certainly the case for our object which experiences a constant acceleration due to gravity. This equation says that the final velocity of some projectile is equal to its initial velocity plus its acceleration times the time elapsed.

Now let’s imagine that we’re looking at the vertical velocity of our object over time. Let’s say that at the outset, at 𝑡 equals zero, the initial velocity of this object in the vertical direction is exactly 19.6 meters per second. If we then wait one second, this equation of motion tells us that the final velocity of our object will be less than its initial value because it’s being slowed down by the acceleration due to gravity. That is, even though it started out moving upwards at a relatively high speed, now, sometime later, it’s moving upward but at a lower speed.

If 𝑔 is exactly negative 9.8 meters per second squared with the negative sign because 𝑔 points downward, then after one second has passed, the final velocity, the velocity in the vertical direction of our object is 19.6 meters per second plus negative 9.8 meters per second squared times one second. This works out to positive 9.8 meters per second. After one second is past, the vertical velocity of our object is one-half what it was at a time of 𝑡 equals zero seconds. Now let’s advance time one more second. So we now have a time of two seconds in our kinematic equation of motion. Our object’s final velocity in the vertical direction is now zero meters per second. That is, it’s stationary.

If we consider the object’s vertical velocity at a time of three seconds, we now have a negative velocity, negative 9.8 meters per second. Lastly, after four seconds had elapsed, the object’s vertical velocity is negative 19.6 meters per second. Over time then, the vertical velocity of this object looks like this. It’s important to note that here we’re talking about velocity, which is a vector quantity. If we were thinking instead of speed, a scalar, then all of the negative velocities would reflect about the horizontal axis. That is, our object’s vertical speed over time will look like this. This settles the question of whether our object’s vertical speed looks like graph (c) or graph (d).

Because this equation of motion is a linear equation, that is, all of the factors involved are to the first power, when we calculate the vertical speed of our object over time, that graph will be linear too. For our answer then, we choose option (c).

Let’s look now at part two of this question.

Which of the graphs (e), (f), (g), and (h) shows the changes in the horizontal speed of the object between leaving the ground and returning to the ground?

We’re now considering how our projectile moves horizontally over time. To see how this works, we can again consider our general equation of motion here. This equation can apply to either vertical or horizontal motion. In the vertical direction, our object experienced a nonzero acceleration, the acceleration due to gravity. In the horizontal direction though, there is no such acceleration. Another way to say that is that mathematically this acceleration equals zero. In that case, this equation takes on a simpler form. The final velocity of our object equals its initial velocity.

Notice that the passage of time has no effect at all on this equation. This means that at any time after our object was launched, so long as it hasn’t reached the ground again, its velocity and really its speed in the horizontal direction will be constant. It will be the same as it was at the outset.

Considering our four graphs, we’re looking for the graph that shows a constant horizontal speed, one whose speed doesn’t change with time. Of all these graphs, only graph (e) meets that condition. This graph shows us a horizontal speed that does not change over time. For our answer then, we choose graph (e). This graph shows the horizontal speed of the object between when it leaves the ground and when it returns to the ground.

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