Topic: Engineering Mechanics
A rocket with lift-off mass of $m_0$ is launched from ground level. During flight, fuel burns at a constant rate for t seconds and exhaust gases are ejecting from the bottom of the rocket at $\beta$ Kg/sec with speed of c $m/s$ relative to the rocket. Ignoring air resistance and assume acceleration due to gravity, g as constant, which of the following expression represents velocity of rocket $v(t)$.
- A. $-c \ln[\frac{m_0-\beta t}{m_0}]-gt$
- B. $-c \ln[\frac{m_0-\beta t}{m_0}]+gt$
- C. $-c \ln[\frac{m_0}{m_0-\beta t}]+gt$
- D. $-c \ln[\frac{\beta}{m_0-\beta t}]-g t$
Answer: A. $-c \ln[\frac{m_0-\beta t}{m_0}]-gt$
Explanation: Using the ideal rocket equation with gravity, the velocity change is given by $v(t) - v(0) = c \ln(m_0 / m(t)) - gt$. Since the fuel burns at a constant rate $\beta$, the mass at time t is $m(t) = m_0 - \beta t$. Assuming initial velocity $v(0) = 0$, $v(t) = c \ln(\frac{m_0}{m_0 - \beta t}) - gt = -c \ln(\frac{m_0 - \beta t}{m_0}) - gt$.