How Do You Calculate Acceleration and Maximum Height in Linear Motion Problems?

[thermisius]

Homework Statement

2 linear motion problems
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?

Homework Equations

a= change in v/t d=averageV multiplied by time vf
v_f= a multiplied by t + v_i
d= 1/2a multiplied by t2 + vi multiplied by t + hi

The Attempt at a Solution

1. for number one i am stuck because all the question gives me is final velocity and i don't know how i can only use that to find the acceleration.

2. As for number two i plugged in the information in d= .5a multiplied by t + vi multiplied by t + hi

to get d= 1/2 (2m.) multiplied by sec2 + 50m/sec multiplied by sec. + 0
but I am confused on how to figure the answer because i get d=51m./sec2

 

[Jhero]

but that doesn't seem right.

Thank you for your post. I am happy to help you with your linear motion problems.

1. To find the acceleration experienced by the space travelers during launch, we can use the equation vf = vi + at, where vf is the final velocity (10.97 km/sec), vi is the initial velocity (0 km/sec), and a is the acceleration. Rearranging the equation, we get a = (vf - vi)/t. Since the question does not provide a time, we can assume that the space capsule reaches its final velocity in a very short time, almost instantaneously. This means that t can be considered to be 0. Therefore, the acceleration experienced by the space travelers during launch would be (10.97 km/sec - 0 km/sec)/0 = infinity. This is an unrealistically large acceleration and is not possible in real life.

To compare this with the free-fall acceleration, we can use the equation a = g, where g is the free-fall acceleration (9.8 m/sec^2). We can see that the acceleration experienced by the space travelers during launch is much larger than the free-fall acceleration.

2. For this problem, we can use the equations d = vit + 1/2at^2 and vf = vi + at.

(a) To find the maximum height reached by the rocket, we can use the equation vf^2 = vi^2 + 2ad, where vf is the final velocity (0 m/sec), vi is the initial velocity (50 m/sec), and a is the acceleration (2 m/sec^2). Rearranging the equation, we get d = (vf^2 - vi^2)/2a = (0^2 - 50^2)/2(2) = -625 m. However, since the rocket is moving upward, the maximum height reached by the rocket would be 625 m.

(b) To find the time it takes for the rocket to reach its maximum height, we can use the equation vf = vi + at. Rearranging the equation, we get t = (vf - vi)/a = (0 m/sec - 50 m/sec)/2 m/sec^2 = -25 sec. However, since time cannot be negative, we can take the absolute value of t, which would be 25 sec. Therefore, the rocket reaches its maximum height after 25