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lina2011 [118]
3 years ago
11

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
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Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

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If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

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\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

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1 year ago
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