Answer to Question #238298 in Physics for JayJay

Question

A rocket rises vertically, from rest, with an acceleration of 3.2m/s² until it runs out o fuel at an altitude of 1200m.

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

(b)How long does it take to reach the point where the rocket runs out of fuel?

(c)What is the maximum height the rocket reaches?

(d)How long does it take to reach max height?

(e)How fast does the rocket hit the ground after falling back to earth?

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

Answer 1

(a) We can find the velocity of the rocket when it runs out of fuel from the kinematic equation:

$$v^2=v_0^2+2ah,$$ $$v=\sqrt{2ah}=\sqrt{2\cdot3.2\ \dfrac{m}{s^2}\cdot1200\ m}=87.64\ \dfrac{m}{s}.$$

(b) We can find the time taken to reach the point where the rocket runs out of fuel from the kinematic equation:

$$v=v_0+at,$$ $$t=\dfrac{v}{a}=\dfrac{87.64\ \dfrac{m}{s}}{3.2\ \dfrac{m}{s^2}}=27.39\ s.$$

(c) Let's first find the height that the rocket reaches after it runs out of fuel:

$$v^2=v_0^2-2gh',$$ $$0=v_0^2-2gh',$$ $$h'=\dfrac{v_0^2}{2g}=\dfrac{(87.64\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=391.87\ m.$$

Finally, we can find the maximum height the rocket reaches:

$$h_{max}=h+h'=1200\ m+391.87\ m=1591.87\ m.$$

(d) Let's first find the time taken from 1200 m to the maximum height:

$$v=v_0-gt',$$ $$0=v_0-gt',$$ $$t'=\dfrac{v_0}{g}=\dfrac{87.64\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=8.94\ s.$$

Finally, we can find the total time to reach the maximum height:

$$t_{tot}=t+t'=27.39\ s+8.94\ s=36.33\ s.$$

(e) We can find the velocity of the rocket when it hit the ground after falling back to earth

from the kinematic equation. Let's take the downwards as the positive direction. Then, we get:

$$v^2=v_0^2+2gh,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot1591.87\ m}=176.64\ \dfrac{m}{s}.$$

(f) Let's first find to reach the ground from the maximum height:

$$h_{max}=\dfrac{1}{2}g(t'')^2,$$ $$t''=\sqrt{\dfrac{2h_{max}}{g}}=\sqrt{\dfrac{2\cdot1591.87\ m}{9.8\ \dfrac{m}{s^2}}}=18\ s.$$

Finally, we can find the flight time of the rocket:

$$t_{flight}=t+t'+t''=27.39\ s+8.94\ s+18\ s=54.33\ s.$$