A rocket moves upward, starting from rest.

[chewy96]

Homework Statement

A rocket moves upward, starting from rest with an acceleration of 32.2 m/s2 for 5.84 s. It runs out of fuel at the end of the 5.84 s but does not stop. How high does it rise above the ground?

Homework Equations

I do not know. All equations I have tried have been missing a variable.

The Attempt at a Solution

I tried finding the velocity and then adding it to what I got when plugging it into xf=xi+vi x t + 1/2g x t^2

 

[hage567]

Homework Statement

A rocket moves upward, starting from rest with an acceleration of 32.2 m/s2 for 5.84 s. It runs out of fuel at the end of the 5.84 s but does not stop. How high does it rise above the ground?

Homework Equations

I do not know. All equations I have tried have been missing a variable.

The Attempt at a Solution

I tried finding the velocity and then adding it to what I got when plugging it into xf=xi+vi x t + 1/2g x t^2

Break the motion of the rocket into two parts. Yes, you need to find the velocity the rocket is at after 5.84 seconds. Then you can use this information in the second stage of its motion.

Show your calculations, so we can see what you are getting.

 

[kerimek]

, but I still could not find the answer.

I would first clarify the given information and make assumptions if necessary. For example, I would assume that the acceleration is constant during the 5.84 s and that the rocket continues to move upward after running out of fuel. I would also assume that the rocket starts from rest at ground level.

Next, I would use the equation vf = vi + at to find the final velocity of the rocket after 5.84 s. Plugging in the given values, we get vf = 32.2 m/s2 x 5.84 s = 187.648 m/s.

Then, using the equation vf^2 = vi^2 + 2ad, we can find the displacement of the rocket (i.e. how high it has risen) by rearranging the equation to d = (vf^2 - vi^2) / 2a. Plugging in the values, we get d = [(187.648 m/s)^2 - (0 m/s)^2] / 2 x 32.2 m/s2 = 571.8 m. Therefore, the rocket has risen 571.8 m above the ground before running out of fuel.

However, this solution assumes that the rocket continues to move upward after running out of fuel. If we assume that the rocket stops moving after running out of fuel, then the displacement would be different. In this case, we would use the equation d = vit + 1/2at^2, where vi = 187.648 m/s, a = -9.8 m/s2 (due to gravity), and t = 5.84 s. Plugging in the values, we get d = (187.648 m/s x 5.84 s) + 1/2 x (-9.8 m/s2) x (5.84 s)^2 = 572.1 m. Therefore, the rocket would have risen 572.1 m above the ground before stopping.

In conclusion, the height the rocket rises above the ground depends on the assumptions we make about its motion after running out of fuel. As scientists, it is important to clarify information and make reasonable assumptions in order to find accurate solutions.