Vectors and Projectiles Legacy Problem #21 Guided Solution

There are two different classes of projectile problems. There's the angle-launched projectile problems, and there's the horizontally-launched projectile problems. This particular problem happens to fall in that second class of a horizontally-launched projectile problem. The reason I know that is I read the problem carefully. I realize that the rock acquires initially a horizontal velocity of 26 meters per second, no vertical velocity. In other words, Dylan launches the rock's sidearm, giving it just horizontal projection with no angle listed, no angle projection. Now, once released from his hand, once projected, the only force acting on it is the force of gravity, which causes it to fall. Originally, there's no y velocity. And then it begins to accelerate vertically and falls a distance of 0.45 meters down at a rate of negative 9.8 meters per second per second. Now, to solve the problem, I recommend that you use the provided x-y table and begin to list known and unknown information. Under the x column of the x-y table, I will list Vox equals 26 meters per second, and I will list Ax equals 0 meters per second per second. Now, that's kind of an implicit information. The fact that it's a projectile means there's no horizontal acceleration. It's 0 meters per second per second. What I'm looking for is the time to fall, and I'm looking for the distance it travels horizontally from the edge of the water, dx. So I'm going to list dx equals question mark. I have two pieces of horizontal information. Under the vertical column, I have three bits of information, one of which is explicitly stated, the fact that this rock falls 0.45 meters to the water. So under that column, I'm going to list dy equals negative 0.45 meters, with that negative meaning it's displaced downwards. I'm also going to list an implied bit of information that Voy is 0. Originally, there's no y velocity. And I'm going to list the thing that I know for all projectiles, Ay equals negative 9.8 meters per second per second. So I know three bits of vertical information. Now, if you go back to the overview page for this set of problems, you're going to see a set of equations, equations for horizontal motion and equations for vertical motion. These two motions are independent of one another, and so their equations can be used separately. We just have to make sure that we don't mix and match vertical information in a horizontal equation. So what I wish to know here as part A is the time it takes to fall. And the value of having the xy table is if you find three bits of information in a column, you know you can use an equation for that dimension of motion. Here we have three bits of vertical information, and so I can use a vertical equation to find the time to fall. So look through your kinematic equations for the vertical motion and find one that has in it Voy and Ay, and use that equation to solve for t. The equation of choice is the one that goes to Dy equal Voy times t plus 1 half Ay t squared. Now, the Voy times t happens to come out to be zero since there's no original y velocity. So the equation simplifies to Dy equal 1 half Ay t squared, where the Ay is negative 9.8 and the Dy is negative 0.45. Plug your values for Dy and Ay into that equation and solve for t. Now, if you do this successfully, you'll get a t value of 0.3030 seconds. Now you need to use that in part B because that question has to do with how far from the edge of the water does the rock travel before it makes its first skip. Find the distance the projectile moves horizontally. Find dx. Now, we only have one horizontal equation of interest, and that is the equation that dx equals Vox times t. We've just calculated t, and we know Vox is 26 meters per second. So what I'm going to do is take this 0.303 from part A and multiply it by 26 in order to get dx. I'm using the equation dx equals Vox times t, and I get about 7.879 meters as the distance it travels horizontally.