Professor Dave here, let's discuss vertical motion. We have familiarized ourselves with position, velocity, and acceleration, as well as the kinematic equations that govern the relationship between these parameters. We also saw how we can apply these to simple horizontal motion, and things will be essentially the same when looking at vertical motion. The only difference is that when examining strictly vertical motion, the acceleration will be the acceleration due to Earth's gravity, which is a constant, -9.8 meters per second squared. This value is unique to the earth since the earth has a particular mass and radius. This value would be different if we were on some other moon or planet, but we humans tend to spend most of our time here on earth, so we can just associate acceleration due to gravity with this value unless otherwise specified. Also with horizontal motion, it is arbitrary which direction we regard as positive or negative, though we often assign the positive direction to the right by convention, but with vertical motion the negative direction will always be down towards Earth, and objects will always accelerate towards the Earth, since that is the main source of gravity we experience every day. We can examine objects that are in free fall, which means they were dropped downwards from a standstill, or we can look at objects that have some initial upward velocity. Either way, acceleration will always be -9.8 meters per second squared when on earth. Let's say for example you drop a rock off of a 100 meter tall cliff. How long will it take to hit the ground and how fast will it be traveling when it lands? To answer this we will need the equation regarding position, because we know that the rock must travel 100 meters in the negative direction. We can plug in -100 here. For the initial velocity that will be 0 because when you drop the rock it is from a complete standstill, only once you let go does gravity cause the rock to accelerate towards the ground, so this whole term can be ignored. Now we plug in the acceleration due to gravity and solve for t. Notice that the negative signs on position and acceleration values will cancel out to give a positive value for time. This is very important, as we have to be careful with signs as we do arithmetic, and this will be a common opportunity for error with many students. And we get around 4.5 seconds for the time elapsed during free fall. Now if we want to know how fast it is going when it hits the ground, we can use this equation to solve for velocity. Again initial velocity is 0 and we have -9.8 meters per second per second, times 4.5 seconds which gives us -44.1 m/s at the time of impact. This value is negative because velocity is a vector and the object is traveling in the negative direction. Now let's say that we take another rock and this time we throw the rock straight up in the air with an initial velocity of 10 m/s. This velocity will immediately begin to decrease because of the negative acceleration due to Earth's gravity. The moment it leaves your hand the velocity will decrease until it becomes 0 and then grows in the negative direction. This means it will travel upwards for some amount of time before falling back down to the bottom where the other rock landed. How long will this rock be traveling before it lands? Well let's do the same calculation that we did before, the only difference here is that instead of an initial velocity of 0 we have an initial velocity of 10 meters per second in the positive direction, or upwards, and the usual acceleration which again will be downwards and therefore negative. We plug in -100 meters for the displacement, remembering that even though the distance this rock will travel is greater than the other one, both rocks start and end in the same place. Then we plug in the other numbers. What we are left with is a little trickier than before because we have both a t and t squared term. This means that we will have to use the quadratic equation. If you have forgotten this from algebra class, here it is as it applies to an equation in ax^2 + bx + c form, we simply take the coefficients of the equation and plug them into this expression and we will get two answers as possible values for x. So we have to take our equation and rearrange it to look like the standard form which involves adding 100 to both sides. There's our a, b, and c. If we plug those in we get the two values for x but the positive answer is the one we are looking for, as it doesn't make sense to discuss negative quantities of time, so we can conclude that it takes 5.65 seconds for this rock to hit the ground, a little over a second longer than the one we simply dropped. How fast will this one be traveling at the point of impact? Let's use this equation again. This time we have to use 10 m/s as the initial velocity and then we plug in our acceleration and the time we just calculated. This will give us negative 45.4 m/s at the time of impact, a little more than the other rock because it fell from a slightly higher position, the peak of its trajectory. So we can see that motion in the vertical direction utilizes the same equations and concepts as horizontal motion, except that the acceleration will always be due to gravity, which is held constant at this value. Let's check comprehension. Thanks for watching, guys. 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