User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Sep 5th 2009, 06:45 AM #1 Junior Member   Join Date: Apr 2009 Posts: 20 Mastering Physics practice question A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9 m/s^2 . The acceleration period lasts for time 10.0 seconds 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 9.8 m/s^2. I really need help with this question, if someone could explain me it that would be greatly appreciated. Thx in advance.   Sep 5th 2009, 06:59 AM #2 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Solidsnake, Use equations of motion: s= (ut) + (1/2)at^2 from rest delete (ut) so S= (1/2)at^2 S = (1/2) x 53.9 x 10^2 S= 26411 m sorry S = 2695 m my mistake Last edited by Paul46; Sep 5th 2009 at 07:07 AM. Reason: mistake   Sep 5th 2009, 07:58 AM #3 Physics Team   Join Date: Feb 2009 Posts: 1,425 You must post your attempt or specify your difficulty. I think this may be that you are not getting the required answer. Here is a hint. First using $\displaystyle s_1 = ut + \frac{1}{2} a t^2$ , find the height reached as long as the fuel is available i.e. 10 secs. Note that u = 0 here. Does the rocket stop travelling upwards after this? No because it has kinetic energy and will continue going up till this is converted into potential energy. This additional height say s2 will have to be added to s1 to get the maximum height where its velocity in the upward direction becomes zero. So use $\displaystyle v^2 = u^2 - 2gs_2$ with v = 0 and here u is not zero but u = final velocity after 10 secs (which i will leave for you to find.) From this find s2 The max height will be thus $\displaystyle H = s_1 + s_2$.   Sep 5th 2009, 11:09 AM #4 Banned   Join Date: Aug 2009 Location: UK Posts: 240 YES!, Physicsquest is totally right i made another mistake, i'm having abit of a blonde day today. Last edited by Paul46; Sep 5th 2009 at 11:14 AM.   Sep 6th 2009, 05:44 AM #5 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Physicsquest, I'm absolutely stumped on finding s2. v^2 = 0 u^2 = 290521 V^2 = u^2 - 2gs2 so if u^2 = 290521 then 2gs2 must equal 290521 so if g =9.8 then 290521/ (2g) = s2 = 14822.5???? What am i doing wrong? My intuition tells me that perhaps V^2 = u^2 - 2gs2^2 which would result in s2 = 121.747 m ? I'm stumped! Last edited by Paul46; Sep 6th 2009 at 06:01 AM.   Sep 6th 2009, 10:51 AM #6 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Physicsquest, Could s2 be 490 m ? this s2 is bugging me please put me out my misery LOL! If it is 490 m i have found the equation. Last edited by Paul46; Sep 6th 2009 at 10:56 AM.   Sep 6th 2009, 11:30 AM #7 Banned   Join Date: Aug 2009 Location: UK Posts: 240 I think s2 = 490 m The equation is the rocket is accelerating for 10 secs so S2= (1/2) (-a)t^2 S1 = 2695 H= S1 - (1/2) (-9.8m/s^2) t2 = 3185 m Last edited by Paul46; Sep 6th 2009 at 11:38 AM.   Sep 6th 2009, 12:16 PM #8 Physics Team   Join Date: Jul 2009 Posts: 310 So the ship is accelerating 53.9m/s^2, and then cuts off the thrusters. The ship is still travelling upwards and accelerating (in some direction), but it's acceleration is going to become g. $\displaystyle s_1 = \frac {1}{2}at^2$ $\displaystyle s_1 = 2695 \mbox { m}$ Now, the speed achieved at this height: $\displaystyle v = at = (53.9)(10) = 539 \mbox { m/s}$ Since acceleration is cut off to the ship, only gravity is acting on it now. $\displaystyle \frac {V_f^2 - V_i^2}{2a} = s_2$ $\displaystyle s_2 = 14822.5 \mbox { m}$ $\displaystyle H = s_1 + s_2 = 2695 + 14822.5$ $\displaystyle H = 17517.5 \mbox { m}$ __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC   Sep 6th 2009, 12:44 PM #9 Banned   Join Date: Aug 2009 Location: UK Posts: 240 I'm not even going to question it Deco. I'm joking when i say this. but i think of John Mcenroe: " You CAN'T BE SERIOUS!!!"   Sep 6th 2009, 02:58 PM #10 Junior Member   Join Date: Sep 2009 Location: berkeley Posts: 19 Maybe I'm beating a dead horse here, but I'll give it a go anyway. This is a two-step problem (as mentioned above), and I'll try to break it down step by step. It's a bit longer, but the equations are more familiar, maybe. Initial givens (for the first ten seconds). This is part 1: $\displaystyle a=53.9$ From this, we can get: $\displaystyle v(t)=53.9t$ $\displaystyle y(t)=\frac{1}{2}53.9t^2$ Now plug in $\displaystyle t=10$ seconds: $\displaystyle y(10)=\frac{1}{2}53.9(10)^2= \boxed{2695(m)}$ $\displaystyle v(10)=53.9*10= \boxed{539(m/s)}$ Now, on to step 2. The givens have changed, we reset our time to 0, because we're already in motion, yadda yadda: $\displaystyle a=-9.8$ $\displaystyle v_0=539$ $\displaystyle y_0=2695$ and can plug these numbers into the equations: $\displaystyle a(t)=-9.8$ $\displaystyle v(t)=at+v_0=-9.8t + 539$ $\displaystyle y(t)=\frac{1}{2}at^2+v_0t+y_0=\frac{1}{2}(-9.8)t^2+539t+2695$ and we need to solve for v(t)=0, because we want to know how long after the fuel runs out that the velocity hits 0 and the rocket starts to fall and ahem: $\displaystyle v(t)=0=-9.8t+539$ $\displaystyle t=55$ and now we can solve y(55) to get our max height: $\displaystyle y(55)=\frac{1}{2}(-9.8)(55)^2 + 539(55) + 2695$ $\displaystyle y(55) = \boxed{17517.5(m)}$  Tags kinematics, mastering, one-dimensional motion, physics, practice, question ,

,

# find the maximum height reached by the rocket ignore air resistance

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post nrahman1 New Users 1 Sep 12th 2016 07:53 AM p75213 Kinematics and Dynamics 0 Jul 4th 2010 09:03 PM Solid8Snake Kinematics and Dynamics 3 Sep 27th 2009 03:04 AM Solid8Snake Kinematics and Dynamics 1 Sep 7th 2009 10:05 PM Heavyarms2050 Advanced Mechanics 5 May 3rd 2008 06:37 PM