Chapter 2 Problem 58

Problem

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s$^{2}$ until it runs out of fuel at an altitude of 950 m. After this point, its acceleration is that of gravity, downward.

(a) What is the velocity of the rocket when it runs out of fuel?

(b) How long does it take to reach this point?

(c) What maximum altitude does the rocket reach?

(d) How much time (total) does it take to reach maximum altitude?

(e) With what velocity does it strike the Earth?

(f) How long (total) is it in the air?

Solution

(a)

Choose upward to be the positive direction, and $y_{0}=0$ at the ground. The rocket has $v_{0}=0$, $a=$ 3.2 m/s$^{2}$ , and $y=$ 950 m when it runs out of fuel.

$v_{\textrm {950m}}^{2}=v_{0}^{2}+2a(y-y_{0})\rightarrow$ $v_{\textrm {950m}}=\pm {\sqrt {v_{0}^{2}+2a(y-y_{0})}}=\pm {\sqrt {0+2(3.2{\textrm {m/s}}^{2})(950m)}}=77.97{\textrm {m/s}}\approx 78{\textrm {m/s}}$

The positive root is chosen since the rocket is moving upwards when it runs out of fuel.

(b)

$v_{\textrm {950m}}=v_{0}+at_{\textrm {950m}}\rightarrow$ $t_{\textrm {950m}}={\frac {v_{\textrm {950m}}-v_{0}}{a}}={\frac {77.97{\textrm {m/s}}}{3.2{\textrm {m/s}}^{2}}}=23.97{\textrm {s}}\approx 24{\textrm {s}}$

(c)

For this part of the problem, the rocket will have an initial velocity $v_{0}$ = 77.97 m/s , an acceleration of $a$ = -9.80 m/s$^{2}$ , and a final velocity of $v$ = 0 at its maximum altitude.

$v^{2}=v_{\textrm {950m}}^{2}+2a(y-950{\textrm {m}})\rightarrow$

$y_{\textrm {max}}=950{\textrm {m}}+{\frac {0-v_{\textrm {950m}}^{2}}{2a}}=950{\textrm {m}}+{\frac {\left(-77.97{\textrm {m/s}}\right)^{2}}{2\left(-9.80{\textrm {m/s}}^{2}\right)}}=950{\textrm {m}}+310{\textrm {m}}=1260{\textrm {m}}$

(d)

The time for the “coasting” portion of the flight:

$v=v_{\textrm {950m}}+at_{\textrm {coast}}\rightarrow$ $t_{\textrm {coast}}={\frac {v-v_{0}}{a}}={\frac {\left(0-77.97{\textrm {m/s}}\right)}{\left(-9.80{\textrm {m/s}}^{2}\right)}}=7.96{\textrm {s}}$

Thus the total time to reach the maximum altitude:

$t=23.97{\textrm {s}}+7.96{\textrm {s}}=32.33{\textrm {s}}\approx 32{\textrm {s}}$

(e)

For the falling motion of the rocket, $v_{0}$ = 0 m/s , $a$ = -9.80 m/s$^{2}$ , and the displacement is -1260 m (it falls from a height of 1260 m to the ground).

$v^{2}=v_{0}^{2}+2a(y-y_{0})\rightarrow$

$v=\pm {\sqrt {v_{0}^{2}+2a(y-y_{0})}}=\pm {\sqrt {0+2(-9.8{\textrm {m/s}}^{2})(-1260{\textrm {m}})}}=-157{\textrm {m/s}}\approx 160{\textrm {m/s}}$

The negative root was chosen because the rocket is moving downward, which is the negative direction.

(f)

The time for the rocket to fall back to the Earth

$v=v_{0}+at\rightarrow$ $t_{\textrm {fall}}={\frac {v-v_{0}}{a}}={\frac {\left(-157{\textrm {m/s}}-0\right)}{\left(-9.80{\textrm {m/s}}^{2}\right)}}=16.0{\textrm {s}}$

Thus the total time for the entire flight:

$t=32.33{\textrm {s}}+16{\textrm {s}}=48.33{\textrm {s}}\approx 48{\textrm {s}}$