View Full Version : Physics on accleration
ryty
Sep 22, 2009, 08:54 AM
A model rocket blasts off and moves upward with an acceleration of 12m/s^2 until it reaches a height of 30m , at which point its engine shuts off and it continues its flight in free fall.
A-what is max height?
B-What is speed just before the rocket hits the ground?
C-What is rockets total duration of flight?
Can someone please explain this to me, and please show your work.
Unknown008
Sep 22, 2009, 09:13 AM
A. Find the final velocity at 30 m height.
This is found using v^2 = u^2+2as
Since u (initial velocity = 0), v^2 = 2as = 2(12)(30)
v = \sqrt{2(12)(30)}
Use again the formula v^2 = u^2+2as.
Now, the final velocity (when the engines are off) is zero, the initial velocity u is \sqrt{2(12)(30)} and the acceleration -9.81 m/s^2.
Solve for s, the distance covered during this ascension.
B. Use again v^2 = u^2+2as.
The initial velocity u is zero (from the highest point), the final velociy v unknown, the acceleration is due to gravity, +9.81 (positive since it is in the same direction as the downwards motion) and the distance s is the max height you just got above in A.
C. Find the time during ascension with motors on. This is given by s = ut + \frac12 at^2 where u = 0, a = +12 and s = 30.
Then, second part, when it goes up with engines off then going down.
v = u+at where v is the final speed of -B (B is the speed you got in part B), u the velocity at 30 m upwards and a = -9.81 m/s^2
I hope it helped! :)
NB: Just in case you didn't know, we aren't supposed to give you the answers directly. I see you had another question. Why not post what you tried, to see where the problem is.
ebaines
Sep 22, 2009, 09:20 AM
I'll be glad to explain the methods you can use, but I won't show my work - this is your homework assignment, not mine! So please try to solve the problem on your own, and if you still get stuck post back and show us how far you've gotten, and we'll help you along.
First thing to recognize is that there are two distinct phases to the flight. When the engine is on the rocket has an acceleratoin of +12 m/s^2, and when the engine is off it has an acceleraton of -9.8 m/s^2 (due to gravity). So you need to figure out the rocket's velocity at the end of the first phase, and then use that figure as the initial velocity for the rocket in the second phase. You can solve everything here with three basic eqations:
v_f = v_i + at
d_f = d_i + v_i*t + 1/2 at^2
v_f^2 - v_i^2 = 2ad
For example, the velocity of the rocket at 30 meters can be found from the formula v^2 = 2ad. And the time it takes to reach 30 meters altitude can be calculated either by using from v = at (using the velocity you just calculated) or from d = 1/2at^2; both approaches give the same answer. See if you can take it from here, and post back is you still are having difficulty.
ryty
Sep 22, 2009, 12:32 PM
u=sqrt2*30*12=26.8m/s
then I divided it by 9.81*2 and got 1.37, then added 30, and got 31.37, and I was wrong
ebaines
Sep 22, 2009, 01:42 PM
u=sqrt2*30*12=26.8m/s
Right - that's the upward velocity at height of 30m.
then i divided it by 9.81*2 and got 1.37,
The formula is v^2 = -2gd (note the velocity term is squared) where v = your initial velocity of this phase as you already calculated it, and g = 9.8m/s^2. Looks like you forgot to square that initial velocity:
(26.8m/s)^2 = 2*9.81m/s^2 * d
d = (26.8m/s)^2/(2 * 9.82 m/s^2) = 36.7m
You should always include the units in your calculations - that way if you make a mistake you will see that the units don't work out.
then added 30, and got 31.37, and i was wrong
Add the above to 30m to get... 66.7m.