You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (b) It is moving back down?

Video transcript

Welcome back everybody. We have a spherical stone that is launched vertically upward. And after some time what goes up must come down now, when it is going down right, we are going to have some sort of vertical acceleration acting downward and we need to find what this is. Well, when the stone is falling there are a couple of forces acting on it. We have a downwards force and an upwards force downwards force is of course the force due to gravity, but the upwards force is a drag force. Now this drag force is proportional to the terminal velocity squared by some proportionality constant, which I'm just going to call C. Now we want to find this vertical acceleration at a speed that is half of the terminal velocity. So what is our A Y. At this speed? Right here, let's use Newton's second law here. Newton's second law states that the sum of all forces in a certain direction, we're gonna look at the Y direction in this case is equal to mass times some acceleration. And this is the acceleration that we are looking for. Now, plugging in our values, we have our drag force minus are forced due to gravity equal to our mass times are vertical acceleration, but it's going to be negative since it is in the negative y direction. Now dividing both sides by negative M. We get that our vertical acceleration is equal to our force due to gravity minus our drag force all over our mass. But what is our drag force? Well, we're told that at terminal velocity a terminal velocity, our drag force is equal to our force to gravity. But we're not looking at terminal velocity. We're looking at this speed right here. Right? So just as a reminder, our drag force which is equal to our force due to gravity at terminal velocity is proportional to our terminal velocity squared. But let's sub in this velocity right here, shall we? So our drag force equal to our proportionality constant times. Are terminal velocity over two squared equal to 0.25 times C squared. Well at terminal velocity C v t squared is just equal to MG. So this is just equal to 0.25 MG. Now let's plug this into our formula four A. Y. We have that A. Y. Is equal to MG minus our drag force equal to 0.25 MG. All divided by our mass Yielding us 0.75GS corresponding to our answer choice of E. Thank you all so much for watching. Hope this video helped. We will see you all in the next one