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
[tex]\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)[/tex]
[tex]\implies y_{\mathrm{max}}=95\,\mathrm m[/tex]
