1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What was the initial velocity of the rocket

  1. Feb 14, 2017 #1
    1. The problem statement, all variables and given/known data
    A fireworks rocket is launched vertically upward, and explodes into two equal-mass pieces at the top of its trajectory. The pieces hit the ground at t1 and t2 seconds after the explosion. What is the rocket’s launch speed? Assume no air resistance.

    2. Relevant equations
    v1 = velocity of mass (t1)
    v2 = velocity of other mass(t2)
    hmax = max height of rocket
    3. The attempt at a solution
    I found the max height by kinematic equations with ma=-mg and then rearranged that height in terms of the initial velocity of the rocket. I set up the energy equation (2m)ghmax = (1/2)mv12+(1/2)mv22. I replaced the velocity of mass 1 with gt1 and of mass 2 with gt2 and solved for the initial velocity. Assuming I have done the math correctly, does this approach make sense? Am I missing a concept?
     
  2. jcsd
  3. Feb 14, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Any clue for us helpers as to what are the given/known data ? You make it look as if only t1 and t2 were given !
     
  4. Feb 14, 2017 #3
    Those are the only givens - that's all you need
     
  5. Feb 14, 2017 #4
    The final answer needs to only be in terms of t1,t2 and g is a known constant so that's fine too. The other variables I represented were ultimately in terms of t1,t2 at the end. But is the physics right?
     
  6. Feb 14, 2017 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see. Why ##v_1 = gt_1## ?
     
  7. Feb 14, 2017 #6
    Because if I solve for the velocity of one of the masses using newtons laws then it comes out to v1 = gt1. The initial velocity is zero because the rocket is at 0 velocity at the moment it splits into two, so I didn't include it. t1 was given at the time for which the mass hits the ground - so I assumed v1 = gt1 is the velocity at the time it hits the ground.
     
  8. Feb 14, 2017 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In that case ##t_1## and ##t_2## would be equal, right ? Both start at the same height and with the same vertical velocity ....
     
  9. Feb 14, 2017 #8
    Yes, lol. Thank you.
     
  10. Feb 14, 2017 #9

    gneill

    User Avatar

    Staff: Mentor

    You found the max height of what, and how? Suppose the explosion launched one piece directly upwards and the other directly downwards with some velocity each. The maximum height of one piece will not be the same as that of the other, and their unknown vertical speeds will complicate things.
    How do you account for the unknown amount of KE introduced by the explosion? Any vertical component added to the speeds of the pieces will affect the timings.

    I think I'd take a closer look at the explosion itself and determine a relationship between the vertical velocity components immediately after the explosion, then write equations for the vertical motion for each.
     
  11. Feb 14, 2017 #10
    I found the max height of the rocket still intact before exploding into two pieces. I have shown that when the pieces explode they are then going in opposite directions - I will work with this. I think I can solve this now, thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted