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(III) Assume a net force $F=-m g-k v^{2}$ acts during theupward vertical motion of a $250-\mathrm{kg}$ rocket, starting atthe moment $(t=0)$ when the fuel has burned out and therocket has an upward speed of 120 $\mathrm{m} / \mathrm{s}$ . Let $k=0.65 \mathrm{kg} / \mathrm{m}$ .Estimate $v$ and $y$ at 1.0 -s intervals for the upward motiononly, and estimate the maximum height reached. Compareto free-flight conditions without air resistance $(k=0)$

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$$730 \mathrm{m}$$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Simon Fraser University

Hope College

University of Winnipeg

Lectures

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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A catapult launches a test…

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14:46

A small 8.00-kg rocket bur…

03:59

(II) $(a)$ Determine a for…

08:38

A small $8.00 \mathrm{~kg}…

04:02

A small 8.00 -kg rocket bu…

05:24

A rocket engine can accele…

05:11

Rocket Flight, A model roc…

10:16

A small $5.00 \mathrm{~kg}…

12:11

A catapult launches a rock…

04:47

A test rocket is fired ve…

05:26

A small $8.00-\mathrm{kg}$…

So we're given f equals two negative mg Linus ky v squared. So force is a function of velocity year and also we know that force is equal to mass times acceleration. So combining these two, we see that exploration is equal to negative G minus key over M V script. So what we did here is we divided am on both sides, toe get rid of em in front of a N g. No for T equals zero. That means that initial time, they're displacement zero and the velocity is given as 1 20 meter per second, which we call V zero, as in initial velocity and acceleration is just the gravitational ex elation. So that's gonna be, um, minus G minus k over M. He's squared, which is negative 9.8 meters per seconds. Quit. So if we assume that this exploration is constant over the next time interval, so then, uh, we can use very secretions like why one equals toe y zero plus B zero identity, plus her off e zero delta t squared for to figure out the displacement. Then for the velocity we can use, the one equals V zero. Plus, is your identity toe. Get the velocity at the next interval on DA. For the circulation, it's just a one which is equal. Do negative G minus K over envies squared where V is V one in our kiss. So if we if we keep on doing so, we'll see that, um, we'll get various results off displacement, velocity and acceleration. And ah, we can see various results off that in this table where the first interval is from 0 to 1. Then we go toe to toe and so on, so forth, and we see that their various velocities and exhalations associate it with this Now, Ah, to reach the maximum height, um, we see that the maximum height is reached when why is equal to 45 meters. So yeah, so that's gonna be the ex elation for us. Sorry. That's gonna be the maximum height for us. And we call it why Max, which is 2 45 meters now. If the air resistance is totally ignored than the ax, elation is a constant negative g. So in that case, we can use the question toe do point of C to get the maximum height. So from this equation, we know that the final velocity squared minus the initial velocity squared is equal to twice the ex elation times the displacement where, uh, in our case, this this last one is the maximum height that can be achieved by dividing the squared by we not scared by two A. What is in our case, uh, is negative, g. So it's gonna be squared minus B squared, minus divided by minus two G and ah, that becomes 1 20 me never second hole squared. So since the final velocity zero So it's 1 20 meters per second. Then we get rid of the minus. Sign here and doing so because we have a minus sign here and we were mind saying in front of er duty. So that becomes 9.8 meters per second squared, ordered of the minor science. And finally this becomes 7 30 meters. Now, as we see here that ah, the maximum due to the air resistance, the maximum height reduces to about 1/3 off the, um ah, 1/3 off the new resistance value. So I'm more so, uh, yeah. So a detailed analysis can give us the maximum height last resort to if we actually decreased the time interval. But if we're taking one second as our time interval then to 45 is okay. And also, um, the spreadsheet used for this problem can be found on the media manager with filing, uh, be s e four. Underscore. I s m on the score ch 05 dot x l s on tab problem 5.104 Thank you.

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