Vectors and Projectiles Legacy Problem #28 Guided Solution

This problem falls into the category of an angle-launched projectile. It is a projectile that's initially launched at an angle to the horizontal. Initially, it has horizontal velocity and vertical velocity. A projectile such as this will rise vertically towards its peak position and then fall from its peak position. In this case, it's a motion that ends when it reaches its original height. Now, the time it takes for that projectile to get up to its peak position is equal to the time it takes for it to come down from its peak position. That will be important as we solve the problem. Equally important will be the fact that when the projectile gets to its peak position, its velocity vertically will momentarily be zero meters per second. Now, the solution begins by taking the original velocity of 8.06 and the angle of launch at 30 degrees and using the so-called Vox-Voy equations in order to calculate the original x-velocity and the original y-velocity. After all, our projectile equations do not have a V0 and a theta in it. Instead, they have a V0x and a V0y. So, what we need to do is to calculate these horizontal and vertical components and use them in these two sets of independent equations that we have for projectiles. So, we use the V0 times the cosine theta to get the V0x. We use the V0 times the sine of theta to get the V0y. Now, I'm going to focus on the vertical motion for a while because that will allow me to calculate information about time. So, I take this V0y. It happens to be 4.03 meters per second. And I say, OK, that V0y is going to change to zero meters per second over the first half of the trajectory of the projectile. That is, when it finally gets to the peak of its trajectory, the V0y will be zero. So, I'm going to use the equation that goes something like this. Vfy equals Voy plus At. And solve for the t that it takes, the time that it takes for that projectile's vertical velocity to change to zero. That's the time it takes to go straight up to the peak of its trajectory. So, if I do that, I'll have the answer to question B. Zero equals 4.03 plus negative 9.8 times t. And when I solve that for t, I get 0.411 seconds. That's the time it takes to go to the peak of the trajectory. And if you want to find the total time, take the time to the peak and double it. So, the 0.411 gets doubled and I get 0.822 seconds. Now, that 0.822 seconds is going to be relatively important because I'm going to need it to calculate part d. In part d, I'm supposed to calculate the horizontal distance that A traveled from Charon to Al when it finally returned to the original height. So, the way I'm going to solve that one is I'm going to use my only x equation that's of importance. And that's the one that goes dx equals vox times t. Now, my vox is just the vo times the cosine of theta. And my t is the 0.822 seconds. Combining those two values, I will get 5.74 meters as the horizontal distance that the egg traveled. What a toss! Now, the final thing that I have to do is to calculate the height of the egg above its release point when it was at the peak of its trajectory. We're talking again about that midpoint of the full parabolic trajectory of our projectile. We're talking about finding dy when the time is 0.411 seconds. So, to find that, I have probably one of my favorite equations to use. It's the one that says dy equals voy plus vfy divided by 2 times the time. Do that again. dy equals voy plus vfy divided by 2 times the time. Now, all of these equations can be found in the overview page. You see the link there at the bottom of this audio help page. So, what you need to do is take the voy of 4.03, add to it the v final y, or the y velocity when it gets to the peak. And that value is 2, or that value is 0. Take that and divide by 2, and what you're finding is the average velocity from the time of release to the time of reaching the peak, the average y velocity. Now, once you've got that value, which is 4.03 divided by 2, multiply by the time it takes to get to that peak position, which is about 0.411 seconds, and you'll have your dy when it gets to the peak.