Vectors and Projectiles Legacy Problem #25 Guided Solution

A good problem solver has a habit of reading the problem carefully, identifying the given information and the unknown information, organizing it, writing it down, and plotting out a strategy to get from the givens to the unknown. Here if we look at this problem, there are two explicitly stated numerical values that I have to extract and equate with some symbols and make typical kinematic equations, which I use for this topic. Those numbers that I'm given are 521 meters as the horizontal distance traveled by a projectile. That's what we commonly call dx. The other number that I have is the ground speed of the airplane, and since the package, our eventual projectile, is on the plane, it too is moving horizontally with a speed of 59.1 meters per second. So that's the vox of a projectile. Now it's a great strategy to organize this information into xy tables, such as the one that you see below this audio help file on this webpage. So in the x column, I'm right now going to write dx equals 521, I'm going to write vox equals 59.1 meters per second, and for any projectile, I will always write ax equals zero meters per second per second. Now that's three bits of horizontal information. If I think about what vertical information is given, there's none. However, there are two implied bits of information that I know about this projectile, this falling package. One of them is that since it's on a plane that's originally moving horizontally across the sky, that the voy is zero meters per second. When it's released from the plane, originally, it has no vertical velocity, voy equals zero. The other thing that I know about all projectiles, this one included, is that ay equals negative 9.8 meters per second per second. So what I'm trying to find is y, or dy, the vertical displacement of this care package. How far does it fall, and to what height must it be dropped in order to land in the pond, 521 meters horizontally from the drop location? Find dy, that's my unknown. So the strategy that I'll employ involves using the three bits of given x information to calculate a time, and then using the time to calculate the dy value. After all, if you think about your dy, your equations for dy, the most common one used is dy equals voy t plus one half ay t squared. I know voy, and I know ay, so if I can just get the t value, I can solve for dy. So to get the t value, I use the x information. Specifically, I use the equation that dx equals vox times t, where the dx is 521 and the vox is 59.1. So I set up that equation as 521 equals 59.1 times t, and I solve for t. Once I get my value of t, I pause a bit, and then I go to my y equations, and I plug t into the equation, dy equals voy t plus one half ay t squared. Since voy is zero, that term cancels from the right side of the equation, and the equation simplifies to dy equals one half times negative 9.8 times your t, which you just calculated, 8.8156 seconds, and squared, and you use that to solve for dy. Good luck.