How high and how fast will a 25 kg rocket fly without air resistance?

AI Thread Summary
A 25 kg rocket generates 400 N of thrust and burns fuel for 40 seconds, leading to a calculated final vertical velocity of 248 m/s and a height of 11,385.7 meters when fuel runs out. The discussion highlights confusion regarding the time it takes for the rocket to reach the ground after fuel depletion, particularly in accounting for gravitational acceleration. Participants clarify that the rocket enters free fall after burnout and suggest calculating the time from maximum height to ground impact. The original poster resolves their confusion with assistance from others. The conversation emphasizes the importance of including gravitational effects in motion calculations.
David Donald
Messages
31
Reaction score
0

Homework Statement


A 25 kg Rocket generates 400 N of thrust. It is designed to burn its fuel for
40 seconds. If launched vertically what will its final vertical velocity be when the fuel runs out? How high can it fly? How long until it reaches the ground? Neglect mass of fuel, and air resistance.

Please check over my work I feel like I may be making a mistake somewhere

Homework Equations


Kinematics Equations

The Attempt at a Solution


First I summed the forces in the Y direction
ΣFy = Ft - Fg = ma
= (400N - 245N)/(25kg) = a = 6.2 m/s^2

Final Vertical Velocity
Vfy = Voy + at
Vfy = 6.2 m/s^2 * 40s = 248 m/s

(Height when fuel runs out)

Yf = Yo + Voy*t + (1/2)a t^2
Yf = (1/2)*(6.2 m/s^2) (40)^2 = 4960 meters

(FINAL HEIGHT)

Vfy = Voy + at
0 = 248 m/s - 9.8m/s^2 t = 25.31 seconds <---- Time of Final Height

Yf = Yo + Voy*t + (1/2)at^2
Yf = 4960m + 248 m/s * 25.31 s = 11385.7 meters

(How Long until it reaches the ground)
I'm confused how to solve this part...
 
Physics news on Phys.org
Everything looks good up to here:
David Donald said:
Yf = Yo + Voy*t + (1/2)at^2
Yf = 4960m + 248 m/s * 25.31 s = 11385.7 meters
What about the (1/2)at2 term in the equation above?

(How Long until it reaches the ground)
I'm confused how to solve this part...
Can you specify in more detail what is confusing your here?
 
I forgot to include the negative acceleration due to gravity right?
i'm not sure how to find the time of max flight...
 
David Donald said:
I forgot to include the negative acceleration due to gravity right?
Right.
i'm not sure how to find the time of max flight...
Not sure what you mean by "max flight". Try to find the time from "burn out" until the rocket hits the ground. During this time the rocket is in free fall.

{EDIT: Or, you can try to find the time from the point of max height until the rocket hits the ground. This will be easier. You already know the time to reach max height.}
 
Nevermind I figured it out, thank you!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top