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# Thread: PHYSICS QUESTION! Need desperate help! REPS

1. ## PHYSICS QUESTION! Need desperate help! REPS

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9m/s2 . The acceleration period lasts for time 10.0s 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 9.80 m/s2 .

I found the rocket gets to 2695 meters by power, but cant figure out how to get the total height it reaches with gravity acting against it.
Any help??
Reps of course

3. Protip OP, I had to do mastering physics and you can find every problem online just copy paste it, i remember this exact problem to just copy paste it and hundreds of explanations will show up

4. Originally Posted by Dawlphinracer
Protip OP, I had to do mastering physics and you can find every problem online just copy paste it, i remember this exact problem to just copy paste it and hundreds of explanations will show up
Cant find anything bro

5. I know one of you phaggots can figure this out!

7. It is a two part problem:

First with power you said you did (but I ain't checking it).

Second part is the distance traveled with a negative accel. of 9.8, so you need to find the velocity when power stops v = vi + at right?) so you can do that (I think the eq you'll need has like vi^ vf^2 and a and dX).

It is algebra man, and you've learned what, 4 equations so far in class?

8. Originally Posted by daniel1337
Cant find anything bro

Mastering physics problem!?
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 29.4 m/s^2. The acceleration period lasts for time 8.00 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s^2.

2 years ago Report Abuse
This answer shows how far the rocket would fall in time 8.00 s starting from zero velocity.

I got it wrong..how do you find what they're looking for
2 years ago

BOTH of those answers are wrong. If there is someone who can actually explain this problem and help me get it correct that would be awesome.
2 years ago

Who's John
Best Answer - Chosen by Voters

U have to solve this in two parts.

First find the distance until the it stops accelerating.

s = ut + 1/2 at^2
s = 0 times 8 + 1/2 (29.4)8^2
s = 940.8 m

Then find the final velocity at that stage

v = u + at
v = 235.2 this will be the initial velocity for the next calculation

v^2 = u^2 +2as
0 = 235.2^2 + 2(-9.8) times s
s = 55319.04 / 19.6
s = 2822.4 metres

total distance = 940.8 + 2822.4 = 29763.2 metres

---------------------------------------------------------

Fuark I rmbr this **** of a problem i'm telling you dude just google and keep looking it and you will find it good luck man

9. Originally Posted by BroganicChemist
It is a two part problem:

First with power you said you did (but I ain't checking it).

Second part is the distance traveled with a negative accel. of 9.8, so you need to find the velocity when power stops v = vi + at right?) so you can do that (I think the eq you'll need has like v and a and dX with some squares in there lol).

It is algebra man, and you've learned what, 4 equations so far in class?
Dude im so lost in this class its an online class so I havent really learned **** the book just kinda jumps around

10. i dont know what any of those abbreviations mean lol my book has like Ax Vx Gx stuff like that

11. can anyone help figure it out?

12. Originally Posted by daniel1337
i dont know what any of those abbreviations mean lol my book has like Ax Vx Gx stuff like that
You are only working in one dimension, the y-plane. The only subscripts used should be "i" for initial velocity and "f" for final. In the solution posted above "u" is the initial velocity AFTER power has ceased. "s" is distance. Not sure why they aren't using "d". It's the basic kinematic equations that can be found at the bottom of this link:

http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm

13. Originally Posted by daniel1337
can anyone help figure it out?
The answer posted above looks correct in its method. Just check the last line where it looks like the two total distances are added incorrectly.

14. vf = vi + a * t
vf = 0 + 53.9 * 10
vf = 539.0m/s
now you have the speed its traveling when it runs out of gas. So now use that as your new vi and figure out what you now have.

vi = 539.0m/s
vf= 0 (at the peak height of its momentum, vi will briefly be 0 before it starts back towards earth)
d = ?
a = -9.8m/s

vf^2 = vi^2 + 2*a*d
therefore d = (vf^2 - vi^2)/2*a
d = (0 - 290521) / 2 * -9.8
d = (-290521)/(-19.6)
d = 14822.5

Sounds semi logical so I'd go with that.

15. ^That's it. And the distance you mentioned in OP is good, too.

Edit: Use the values in YOUR problem. But the equations/method posted above are legit.

16. Originally Posted by daniel1337
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9m/s2 . The acceleration period lasts for time 10.0s 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 9.80 m/s2 .

I found the rocket gets to 2695 meters by power, but cant figure out how to get the total height it reaches with gravity acting against it.
Any help??
Reps of course
Lol not sherrif serious. It's constant acceleration and one-dimensional vectors ffs...

y=y_i + v_i*t + 0.5at^2

Either memorize this formula, or know the basics (Acceleration = Chng. Velocity / Time; Avg. Velocity = Distance / Time; and you can derive the equations from there if you know partial differentiation)

Since the rocket starts at rest, y_i and v_i are both zero, therefore 0.5*44.1*100 = 2205 meters is when it is out of gas

However, since it is still going even when it is out (it will be decelerating), and the instantaneous velocity at y=2205 is 441 m/s (accel * time), it will still go on for 45 seconds before stopping (and turning around). The distance before it stops will then be y=9922.5, but make sure you add the previous y and you get 12127.5 as max height.

Might want to check answers with someone else though, I rushed through this pretty quick.

17. Originally Posted by Kevlar14
The answer posted above looks correct in its method. Just check the last line where it looks like the two total distances are added incorrectly.
Is there any way you guys can explain to me wtf he just did?!

Lol not sherrif serious. It's constant acceleration and one-dimensional vectors ffs...

y=y_i + v_i*t + 0.5at^2

Either memorize this formula, or know the basics (Acceleration = Chng. Velocity / Time; Avg. Velocity = Distance / Time; and you can derive the equations from there if you know partial differentiation)

Since the rocket starts at rest, y_i and v_i are both zero, therefore 0.5*44.1*100 = 2205 meters is when it is out of gas

However, since it is still going even when it is out (it will be decelerating), and the instantaneous velocity at y=2205 is 441 m/s (accel * time), it will still go on for 45 seconds before stopping (and turning around). The distance before it stops will then be y=9922.5, but make sure you add the previous y and you get 12127.5 as max height.

Might want to check answers with someone else though, I rushed through this pretty quick.
I think your number for the height after the gas runs out is wrong because the masteringphysics thing said I had that part right.... Are you sure you used the right number for acceleration?

19. anyone?

20. ---------- height after acceleration--------------

x(t) = x_0 + v_0 * t + 1/2 * a * t² with x_0 = 0, v_0 = 0, a = 53,9 - 9,8 and t = 10

=> x (10s) = 1/2 * 44,1 * 10² = 2205m

---------- speed after acceleration-------------

v (t) = a * t = 44,1 * 10 = 441m/s

---------- time till zero velocity --------------

v (t) = 0 = v_0` - a * t` = 441 - 9,8*10

=> t = 45s

---------- distance travelled after initial acceleration------

x (t) = x_0` + v_0`*t` - 1/2 * a * t`²

x (45) = 2205 + 441*45 - 1/2 * 9,8 * 45² =2205 + 19845 - 9922,5 = 12127,5m

21. Super easy problem, you definitely shouldn't have taken physics online, it's going to get much much more difficult, so you're going to be in some serious trouble if you're struggling with this. Not trying to be a dick or anything, just being honest. PM me in the future if you have any problems, I just got done with both physics I and II last semester, I may remember enough to help.

22. Originally Posted by DCutch
Super easy problem, you definitely shouldn't have taken physics online, it's going to get much much more difficult, so you're going to be in some serious trouble if you're struggling with this. Not trying to be a dick or anything, just being honest. PM me in the future if you have any problems, I just got done with both physics I and II last semester, I may remember enough to help.
dude trust me I didnt have an option it was only available online. Strong poverty college huh? Ill keep you in my PMs thanks brah

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