# Homework Help: Linear motion questions

1. Sep 2, 2008

### thermisius

1. The problem statement, all variables and given/known data

2 linear motion problems

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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?

2. Relevant equations

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

3. The attempt at a solution
1. for number one i am stuck because all the question gives me is final velocity and i dont 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 im confused on how to figure the answer because i get d=51m./sec2

2. Sep 2, 2008

### thermisius

oops put this under wrong category sry about that

3. Sep 2, 2008

### HallsofIvy

You know the final velocity (vf= 10.97 km/sec), the initial velocity (vi= 0 km/sec), initial height (h= 0 km) and distance traveled (d= 220 m= .22 km). Use d= 1/2 a t2+ vit+ hi and vf= at. You have two equations for the two unknown values a and t.

[/quote]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 im confused on how to figure the answer because i get d=51m./sec2[/QUOTE]
The rocket accelerates at a= 2 m/s2 until it is 150 m high. That means
150= (1/2)(2)t2+ 50t Solve that for t to determine how long the rocket engine fires and then use vf= (2)t+ 50 to determine its speed at that time. After the rocket engine has stopped firing, its acceleration is that of gravity, -9.8 m/s2. Use d= (-9.8/2)t2+ vft+ 150 (vf is the speed after the rocket engine stopped firing that you just found) for its height after that. You can find its maximum value by completing the square.

Last edited by a moderator: Sep 2, 2008