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What was the initial velocity of the rocket

  • Thread starter Vitani11
  • Start date
275
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1. Homework Statement
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. Homework 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?
 

BvU

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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 !
 
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Those are the only givens - that's all you need
 
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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?
 

BvU

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I see. Why ##v_1 = gt_1## ?
 
275
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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.
 

BvU

Science Advisor
Homework Helper
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In that case ##t_1## and ##t_2## would be equal, right ? Both start at the same height and with the same vertical velocity ....
 
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Yes, lol. Thank you.
 

gneill

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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.
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.
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?
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.
 
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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
 

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