Question

A test rocket is launched vertically from ground level (y=0), at time t=0.0s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has to 49m and acquired velocity of 30m/s. What is the maximum height that the rocket will reach
A. 95
B. 80
C.103
D.112
E.184

Answer 1

Once the engine burns out, the rocket is in freefall with initial height 49 m and initial velocity 30 m/s. At its maximum height, its velocity will be 0 m/s, as it's uniformly accelerating downward at 9.81 m/s^2. At this point, we have

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(y_{\mathrm{max}}-49\,\mathrm m\right)$

$\implies y_{\mathrm{max}}=95\,\mathrm m$