What Is the Unrealistic Acceleration in Jules Verne's Moon Launch Scenario?

[thermisius]

hi i have 2 questions on linear motion that i could use some help on step by step.

1. Jules Verne in 1865 proposed sending men to the moon by firing a space capsule from a 220-m cannon with final velocity of 10.97 km/sec.. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration,9.8 m/sec.


2. A model rocket is launched straight upward with an initial speed of 50.0 m/sec.. It accelerates with a constant upward acceleration of 2.00m/sec² until its engines stop at an altitude of 150 m. (a) what is the maximum height reached by the rocket? (b) how long after lift off does the rocket reach its maximum height? (c) How long is the rocket in the air?

 

[Defennder]

1. Use the definition of acceleration here. Assume average acceleration.

2. You need to use the 3 famous kinematics equations here. As for (c), the amount of time it is in the air is the total time taken to go up and for it to come crashing down.

 

[baba89]

1. To calculate the acceleration experienced by the space travelers during launch, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, v = 10.97 km/sec, u = 0, and s = 220 m. Plugging in these values, we get 10.97^2 = 0^2 + 2a(220), which simplifies to 120.2 = 440a. Solving for a, we get a = 0.273 m/sec^2. This is the acceleration experienced by the space travelers during launch.

Comparing this to the free-fall acceleration of 9.8 m/sec^2, we can see that the acceleration experienced during launch is much smaller. This is because the space capsule is being launched with a controlled force and trajectory, whereas free-fall acceleration is due to the force of gravity.

2. (a) To find the maximum height reached by the rocket, we can use the equation v^2 = u^2 + 2as again. In this case, v = 0, u = 50 m/sec, and a = 2 m/sec^2. We also know that s = 150 m, since that is the altitude at which the engines stop. Plugging in these values, we get 0^2 = 50^2 + 2(2)(150), which simplifies to s = 625 m. Therefore, the maximum height reached by the rocket is 625 m.

(b) To find how long after lift off the rocket reaches its maximum height, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, v = 0, u = 50 m/sec, and a = 2 m/sec^2. Plugging in these values, we get 0 = 50 + 2t, which simplifies to t = 25 seconds. Therefore, the rocket reaches its maximum height 25 seconds after lift off.

(c) To find how long the rocket is in the air, we can use the equation s = ut + 1/2 at^2, where s is the displacement, u is the initial