Question

A rocket moves upward, starting from rest with an acceleration of +34.1 for 4.96 s. It runs out of fuel at the end of the 4.96 s but does not stop. How high does it rise above the ground?

Answer 1

"It runs out of fuel at the end of the 4.96 s but does not stop" - if it doesn't stop, then wouldn't it just keep rising? Probably the question should be a bit clearer about the fact that the rocket keeps moving, but after the first 4.96 seconds its movement is due to the downward pull of gravity.

If that's the case, then for the time before fuel runs out, the rocket's upward trajectory is given by

$y=\dfrac12\left(34.1\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

so after 4.96 seconds, it will have risen to a height of

$y=\dfrac12\left(34.1\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.96\,\mathrm s)^2=419\,\mathrm m$

(assuming the given acceleration is indeed in m/s^2). Meanwhile, the rocket will have an upward velocity of

$v=\left(34.1\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.96\,\mathrm s)=169\,\dfrac{\mathrm m}{\mathrm s}$

Then after the fuel runs out, acceleration due to gravity kicks in and the rocket's altitude at time $t$ is given by

$y=419\,\mathrm m+\left(169\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

We could find the maximum value for this quadratic, or we can resort to a simpler method via the formula

$v^2-{v_0}^2=2a(y-y_0)$

At its maximum height $y$, the rocket will have vertical velocity $v=0$. The initial velocity and height $v_0$ and $y_0$, respectively, refer to the velocity and height after 4.96 seconds, so we have

$-\left(169\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)(y-419\,\mathrm m)\implies y=1880\,\mathrm m$