Rocket Height (Motion in streight line)

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[SOLVED] Rocket Height (Motion in streight line)

Hey I'm 75% sure i have the correct answer, but i have to turn it in tomorrow and it's a lot of points so i just want to make sure I'm going it right.

Homework Statement

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 49.0 . The acceleration period lasts for time 9.00 until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 m/s^2 .

Homework Equations

Sorry if this doesn't look so hot, the formulas were given to me with lost of sub things and i don't know how to make them look good on forums, I'm trying the Latex for the first time

y(t)= y$$_{i0}$$ + v$$_{i0}$$t + (at$$^{2}$$)/2
v(t)=v$$_{i0}$$ + at

The i0 should be at the bottom and it means initial at time zero but you probably already knew that.

The Attempt at a Solution

y(t)= y$$_{i0}$$ + v$$_{i0}$$t + (at$$^{2}$$)/2 for 0 < t < 9
Declare variables and set them equal to values in the problem.
y$$_{i0}$$ = 0
v$$_{i0}$$ = 0
a = 49
formula for postion after 9 seconds of acceleration.
y(9)=(49(9)$$^{2}$$)/2

solve to a number = 1984.5 m at t = 9

Now calculate the velocity at time = 9.
v(t)=v$$_{i0}$$ + at
v(9)=0 + 49(9)
v(9) = 441 m/sCreate a new position fuctionfor time after 9 to the time where the rocket is at 1984.5.

y(t)= y$$_{i0}$$ + v$$_{i0}$$t + (at$$^{2}$$)/2 for 9 < t
Declate Variables
y$$_{i0}$$ = 1984.5m
v$$_{i0}$$ = 441 m/s
a = -9.8
t = (t-9) (offset time to account for the first part)

y(t)= 1984.5 + 441(t-9) + (-9.8(t-9)$$^{2}$$)/2

I don't know how to do this properly, but i know if i sub in an x for (t-9) and solve for the x intercepts and then add them and divide by two i will get the time that it's at the top, and then plug that time back into the function and i should get the answer. to avoid the quadratic forumla I'm putting it into grapher on my mac.

and get something around 3*10^5 ?

 

[hage567]

What is the velocity of the rocket at maximum height? Using this information you can work out the time the rocket takes to go from when the fuel runs out to when it reaches maximum height. Then you can find the distance it goes in this time.

 

[Moetasim]

Hi there,

It looks like you have the right idea for solving this problem. You correctly used the kinematic equations to find the maximum height reached by the rocket. Your initial position and velocity are both zero, and you used the acceleration of 49 m/s^2 for the first 9 seconds of the rocket's motion. After that, the rocket is in free fall and experiences an acceleration of -9.8 m/s^2 due to gravity.

You correctly found the maximum height to be 1984.5 meters and the velocity at that point to be 441 m/s. To find the time it takes for the rocket to reach this height, you can use the quadratic formula or graph the equation as you mentioned. Both methods should give you a time of approximately 3 seconds after the fuel is exhausted.

Overall, it seems like you have a good understanding of the problem and have solved it correctly. Just make sure to double check your calculations and units to ensure accuracy. Good luck on your assignment!