How Long Does It Take for a Rocket to Return After Stopping Its Initial Ascent?

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SUMMARY

The discussion focuses on calculating the total time it takes for a model rocket, which accelerates upward at 3.0 g for 15 seconds, to return to the ground after stopping its ascent. The solution involves breaking the problem into two phases: the upward acceleration phase and the subsequent descent phase. Key equations used include the kinematic equations for constant acceleration, specifically ∆y = 1/2 aτ^2 for the ascent and ∆y = v₀t + 1/2at² for the descent. The final answer requires solving for the time taken during the descent after the rocket reaches its peak height.

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  • Understanding of kinematic equations for constant acceleration
  • Basic knowledge of physics concepts such as velocity and acceleration
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Homework Statement


A model rocket takes off from ground level accelerating upward at a = 3.0 g. This upward acceleration lasts for a time τ = 15 s. Afterward the rocket continues upward, eventually stops rising, then falls back to the ground.
How much time passes from the initial upward acceleration stopping to the rocket returning to the ground?

Homework Equations


constant acceleration

The Attempt at a Solution


i broke them down to two different moments (1. during the blast off, 2. when it comes back down.)

moment 1: let t = τ , vf = aτ , ∆y = 1/2 aτ^2

moment 2 : 1) ∆y = volt + 1/2at^2
aτ^2 = aτ - 1/2gt^2
 
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Physicsnoob90 said:

Homework Statement


A model rocket takes off from ground level accelerating upward at a = 3.0 g. This upward acceleration lasts for a time τ = 15 s. Afterward the rocket continues upward, eventually stops rising, then falls back to the ground.
How much time passes from the initial upward acceleration stopping to the rocket returning to the ground?

Homework Equations


constant acceleration

The Attempt at a Solution


i broke them down to two different moments (1. during the blast off, 2. when it comes back down.)

moment 1: let t = τ , vf = aτ , ∆y = 1/2 aτ^2

moment 2 : 1) ∆y = volt + 1/2at^2
aτ^2 = aτ - 1/2gt^2
And?

The "attempt at a solution" means, you know, actually working out the solution, not just throwing a few equations out at random.
 

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