Vectors and Projectiles Legacy Problem #30 Guided Solution

This problem falls into the category of a so-called angle launch projectile problem in which the projectile begins its trajectory initially with both horizontal and vertical velocity. You can always recognize one of these problems because they always tend to give you an angle value in the problem. So here we're told a ball is hit with a speed of 18.7 meters per second at an angle of 65.1 degrees. It would rise up towards its peak of its trajectory. At the peak of its trajectory momentarily its y-velocity would be zero and then it would fall back down from the peak. The time that it would take the projectile to rise to this peak would be equal to the time it takes to fall from its peak. Thus the total time in the air is going to be twice the time it takes to get to the peak. A successful solution to the problem always begins by taking the original velocity V0 and the angle and using the so-called Vox-y equations to find the Vox and the Voy. After all we have no equations that have within them V0 and theta. So we need to find the Vox and Voy and then use our two sets of independent x and y equations in order to solve for the unknown. So I'm going to do that. I'm going to use the Vox-y equations to calculate Vox which is 18.7 times the cosine of 65.1 degrees. I would write that number down and label it Vox and write it down to several decimal places. Then you're going to do the same thing to calculate Voy or at least a similar thing. You're going to go V0 of 18.7 times the sine of 65.1 degrees and that is equal to Voy. Write that number down and you're going to use it a couple of times in your calculations. So I have Vox Voy and now what I'd like to do is find the time it takes to go up to the peak and then double it to get the time to come back down. Finding the time it takes to go to the peak would utilize one of our y equations. So I'm going to look at my list of y equations. You'll find them on the sets overview page. The one I like to use to calculate the time to the peak is the one that goes something like this. VFy equals Voy plus Ay times t. Now the VFy is zero at the peak and the Voy we just calculated using Vox-y equations. I wrote down 16.9617 and the Ay is negative 9.8. So using that equation I go something like this. I say zero equals 16.9617 plus negative 9.8 times t. And I solve for t and that gives me the t it takes to go from the original launch position to the peak position. It comes out to be 1.7308 seconds. Now the total time in the year is simply the double of that time. After all we're just finding the time to go to the halfway point of the trajectory. So I double my number and I get 3.4616 seconds. I hope that you've written both of these numbers down and labeled them as t up or t halfway for 1.7308 seconds and t total for the total time for the trajectory 3.4616 seconds. Now I'm going to find the dy value at the peak of its trajectory. That is how high up, the answer to part b, how high up does this tennis ball go? And so to do that I need some sort of dy equation. There are several equations with dy in it. The one I like to use for this calculation is the one that goes something like this. And then there's a fraction in the numerator. It says voy plus vfy in the denominator. There's a 2 and then multiplied by the time. So I'm going to use that equation to find the distance it rises to its peak. So I say dy equal in the numerator. I'm going to use 16.9617. That's my original y velocity plus 0. That's my final y velocity that is at the peak. Then I'm going to divide by 2. That gives me my average velocity heading up to the peak. Then I'm going to multiply by the time it takes to get to the peak, which I previously calculated. It's 1.7308 seconds. Now all those numbers should be written down for you. You've been following these directions and when you finally finish the calculation, you should get 14.6786 meters. Now the last calculation is part c. Determine the distance the ball travels horizontally before it finally lands. And so that means we need to calculate dx for a time of 3.4616 seconds, the total time of flight. Now fortunately there's only one dx equation that we typically use, and it's the one that goes dx equal vox times t. The vox we calculated from our voxel equations. It was 7.8734. And the time we calculated is the answer to part a, and it was 3.4616. Multiply those two numbers together and you get about 27.2544, or round it to 27.3 meters.