Question

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30 m/s How long did the burn phase last? A: 2.6 s B: 2.3 s C: 3.3 s D:3.6 s E: 2.9 s

Answer 1

The rocket starts at rest, so its initial velocity is 0. If the burn phase lasts $t$ seconds, then during this interval the rocket's velocity is

$v=at\implies a=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}t$

and its position is

$x=\dfrac12at^2\implies a=\dfrac{2(49\,\mathrm m)}{t^2}$

So we have

$\dfrac{30\,\frac{\mathrm m}{\mathrm s}}t=\dfrac{2(49\,\mathrm m)}{t^2}\implies t=3.3\,\mathrm s$

and the answer is C.