# rocket physics problem

;].OW��K���� j%�"�t*s��$�RO�,9[�� ~�I�� �8��NN�����\M�(�������q"%�#D� T�4x R = 22 \text{ kg/s}. 2 0 obj There’s only one problem: It’s going 17,000 miles an hour. It does. Shipment delivered to drop point but I didn't order anything. To determine the change of velocity, use the rocket equation Equation \ref{9.38}. In effect, the efficiency (how much change in energy it produces) of the rocket is changing, depending on when it is used. The initial velocity for this motion is 80 m/s. 10 0 obj With this sort of problem it is often worth sketching a velocity against time graph. In rocket problems, the most common questions are finding the change of velocity due to burning some amount of fuel for some amount of time; or to determine the acceleration that results from burning fuel. (a) How long was the rocket above the ground? The exhaust velocity of fuel relative to the rocket engine is ve=200 m/s. New user? endobj If the rocket is used when the ship is traveling quickly ($v$is large), then it works more efficiently to slow the ship down. <> 5B�v1:�����8���KE��e�Zs�&�#"����? It only takes a minute to sign up. A rocket of mass 5000 kg 5000 \text{ kg} 5000 kg ejects fuel at a rate of R=5 kg/s R = 5 \text{ kg/s} R=5 kg/s with an exhaust velocity of ve=3000 m/s. Its engines then fire and it is accelerated at until it reaches an altitude of 1000 m. At that point the engines fail and the rocket goes into free-fall. Burning earlier makes your flight longer and thus increases your impact speed. endobj How can I successfully compile$\mathbb{5}$? Rocket physics on Brilliant, the largest community of math and science problem solvers. <>>> The acceleration is pointing downward. <> Calculating Rocket Engine Power from Force and Burntime, Dystopian future with telepathic children. 4000 \text{ m/s}. <> M��-���E�_��L�,��Ի>�%E^.��5ʵb����)�׺�K���BN = So the fuel was burned, it's chemical energy is released. And the kinetic energy of the rocket is minimized. endobj Here is a paraphrase of H&R’s derivation. stream ! rev 2020.10.14.37815, The best answers are voted up and rise to the top, Physics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Understanding rocket problem intuitively [closed], Responding to the Lavender Letter and commitments moving forward. endobj The initial velocity is 120 m/s pointing upward. So the trick is to get the rocket to fall as far as possible when its thruster is on as is shown in the diagram below. You're right that the total energy will always be conserved. 11 0 obj Will a rocket produce more thrust if fired in air, rather than vacuum? That is to increase the kinetic energy of outgoing gas as much as possible. Therefore, the most reasonable height to start using the thrust is when the height at which the rocket would end using its thrust is when it reaches the ground (the calculation to find the actual value of the height is very complicated, so I'll skip (it's not the main point of my question). A rocket whose mass is Mi=2600 kg, M_i = 2600 \text{ kg}, Mi​=2600 kg, of which 2080 kg 2080 \text{ kg} 2080 kg is fuel, consumes fuel at a rate of R=26 kg/s. We want to make$E_3$as small as possible (making it negative would be nice: that would mean that kinetic energy of the rocket has decreased). <>/Font<>/XObject<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 960 540] /Contents 10 0 R/Group<>/Tabs/S/StructParents 1>> As far as I see this, the thrusters provide a thrust equal to the weight of the rocket, so firing them can only make the nett force on the rocket zero for$2$seconds, which means the rocket will not accelerate for$2\$ seconds. Then.

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