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Estudos Gerais04/04/2025

Movimento Vertical com Aceleração Linear

Movimento Vertical com Aceleração Linear

F=bj^ \mathbf{F} = -b \hat{j} p=mgj^ \mathbf{p} = mg \hat{j} F=md2ydt2j^ \sum \mathbf{F} = m \frac{d^{2}y}{dt^{2}} \hat{j} (mgbv)=mdvdt (mg - b v) = m \frac{dv}{dt} dydt=gbmv \frac{dy}{dt} = g - \frac{b}{m} v

Quando atingimos o vetor dvdt=0\frac{dv}{dt} = 0

v=0 v = 0 g=limt(θ0θ)=θ0g g = \lim_{t \to \infty} \left( \frac{\theta_{0}}{\theta} \right) = \frac{\theta_{0}}{g} u=u0ebmkt u = u_{0} e^{\frac{b}{m} k t} u=u0etτ u = u_{0} e^{-\frac{t}{\tau}} τ=mb \tau = \frac{m}{b} b=θ0cm+θ0(1etτ) b = \theta_{0} c_{m} + \theta_{0} (1 - e^{-\frac{t}{\tau}})

Resolver a questão de física

Movimento Vertical com Aceleração Linear

\[ \mathbf{F} = -b \hat{j} \]
\[ \mathbf{p} = mg \hat{j} \]
\[ \sum \mathbf{F} = m \frac{d^{2}y}{dt^{2}} \hat{j} \]
\[ (mg - b v) = m \frac{dv}{dt} \]
\[ \frac{dy}{dt} = g - \frac{b}{m} v \]
Quando atingimos o vetor \( \frac{dv}{dt} = 0 \)
\[ v = 0 \]
\[ g = \lim_{t \to \infty} \left( \frac{\theta_{0}}{\theta} \right) = \frac{\theta_{0}}{g} \]
\[ u = u_{0} e^{\frac{b}{m} k t} \]
\[ u = u_{0} e^{-\frac{t}{\tau}} \]
\[ \tau = \frac{m}{b} \]
\[ b = \theta_{0} c_{m} + \theta_{0} (1 - e^{-\frac{t}{\tau}}) \]
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