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Ch 06: Dynamics I: Motion Along a Line

Chapter 6, Problem 6

a. A rocket of mass m is launched straight up with thrust Fₜₕᵣᵤₛₜ. Find an expression for the rocket's speed at height h if air resistance is neglected.

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Hey, everyone in this problem, we're told that a missile encounters four forces during a purely vertical flight, gravity propulsion and aerodynamic forces of lift and drag. We're asked to use equations that model the launch of a missile find an expression for the magnitude of the missile velocity at an altitude of two H. If aerodynamic forces are ignored, we're given four answer choices A through D and they're all different expressions for the velocity V with the variables H the thrust force ft the mass M and the acceleration due to gravity G. Now we're looking for the magnitude of the missile velocity. OK. So we're looking for velocity, we have an altitude, we want to think about our U AM or kinematic equations. Let's go ahead and write down the variables we know. OK. And this is a vertical flight. We're looking in the y direction. So the initial velocity V is going to be 0 m per second. It begins from rest as it launches. OK. So it starts at rest and launches upwards with some acceleration. Now the final velocity VF we're gonna call this V and this is what we're looking for. Yeah, this is a value we're looking, where are we looking for that value or we're looking for that value when the rocket travels a distance of two H has an altitude of two H. And so delta Y that displacement is going to be two H. Now, the acceleration we don't know and the time we don't know either. So right now we have two known values, one that we're trying to find and that is not enough and we need to have three values that we know. So we are given some information about forces in this problem. OK? We know that the forces are related to the acceleration through Newton's second law. So let's go ahead and draw a free body diagram and think about the forces in this problem and see if that helps us get to an acceleration. Do we have a rocket ship? And there's gonna be a thrust force acting upwards that propels this rocket up? And we know that we're also gonna have the force of Grabby acting duck. So our net force is going to be acting up and we know that because our missile is going to reach an altitude of two H. So it's moving upwards, there's a net force going upwards, courses is going to be equal to the mass multiplied by the acceleration. OK? That's Newton's second law. So our net force and we're saying that up is our positive direction, that direction of the net force. This is gonna be equal to the thrust force ft minus the force of gravity. Here, the thrust force is acting in the positive direction. The force of gravity is acting in the negative direction. This is gonna be equal to the mass M multiplied by the acceleration A. Now the force of gravity I recall is going to be the mass multiplied by the acceleration due to gravity. So we can write our equation as FTMM multiplied by G is equal to M multiplied by A. Now we wanna find the acceleration A and so the acceleration A dividing both sides by M is gonna be the thrust force FT minus M multiplied by G all divided by M. We can simplify a little bit and write this as FT divided by MS minus the acceleration due to gravity G. OK. So we have our acceleration now in terms of the other variables and we can use that in our kinematic or U AM equation. So now we have information about VNA delta Y and the acceleration A three known values we're looking for VF OK. So we're gonna choose the kinematic equation with those four variables. That's gonna be VF squared is equal to V not squared plus two A delta substituting in our values. VF is just our velocity that we're looking for V squared. This is gonna be equal to our initial velocity which is zero. OK. So that term goes away multiplied by two or sorry plus two multiplied by the acceleration ft divided by M minus the acceleration due to graph eg multiplied by the displacement delta Y which is two H. All right. So we have two multiplied by two. At the end, we can simplify this expression and write it as V squared is equal to. And we're gonna move the H to the front. OK. And all of the answer choices, the H is before our bracket here with the first force in the other terms. So we're just gonna move it to the front. They're all being multiplied together. So the order does not matter. We can write this as 4h multiplied by the thrust force FT divided by M. Why is the acceleration due to gravity? G? The sulfur V? We're gonna take the square root that B is equal to the square root of 4h multiplied by FT divided by MS minus the acceleration due to gravity G. And we took the positive route. Um We're asked to find the magnitude of the velocity. So we just wanna know that positive value, we don't care about the direction. OK. So we're taking the magnitude, we're taking the positive. All right. Now, we can simplify this a little bit. We know that the square root of four is equal to two. And so we can take that square root, leave everything else under the root and write this as the V is equal to two, multiplied by the square root of H multiplied by thrust force ft divided by M minus the acceleration due to gravity G. And that is the final expression for the magnitude of the missile's velocity at an altitude of two H. If we compare this to the answer choices we weren't given. And we can see that this corresponds with C, the answer choice has just written the square root as the exponent one cap and those two are equivalent. And so C is the correct answer here. Thanks everyone for watching. I hope this video helped see you in the next one.