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 .
Write your answer numerically in units of meters.
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??
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09082013, 01:19 AM #1
PHYSICS QUESTION! Need desperate help! REPS
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09082013, 01:21 AM #2

09082013, 01:21 AM #3

09082013, 01:22 AM #4

09082013, 01:24 AM #5

09082013, 01:26 AM #6

09082013, 01:26 AM #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?

09082013, 01:26 AM #8
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.
Write your answer numerically in units of meters.
2 years ago Report Abuse
Additional Details
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

09082013, 01:28 AM #9

09082013, 01:30 AM #10

09082013, 01:32 AM #11

09082013, 01:32 AM #12
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You are only working in one dimension, the yplane. 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

09082013, 01:34 AM #13

09082013, 01:35 AM #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
add your 2695 and you get 17518m.
Sounds semi logical so I'd go with that.

09082013, 01:37 AM #15

09082013, 01:39 AM #16
Lol not sherrif serious. It's constant acceleration and onedimensional 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.Last edited by Addyman; 09082013 at 01:47 AM.
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09082013, 02:03 AM #17

09082013, 02:08 AM #18

09082013, 02:12 AM #19

09082013, 02:17 AM #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 = 12127,5mLast edited by Xeledon; 09082013 at 02:30 AM.
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09082013, 02:26 AM #21
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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.
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09082013, 02:34 AM #22
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