Answer
(a) The velocity of the rocket when it runs out of fuel is 70 m/s.
(b) It takes 22 seconds until the rocket runs out of fuel.
(c) The maximum altitude is 1025 m.
(d) The total time to reach maximum altitude is 29 s.
(e) The rocket strikes the earth with a velocity of 140 m/s.
(f) The total time in the air is 43 s.
Work Step by Step
(a) $v^2 = v_0^2 + 2ay$
$v = \sqrt{0 + (2)(3.2 ~m/s^2)(775 ~m)} = 70 ~m/s$
The velocity of the rocket when it runs out of fuel is 70 m/s.
(b) $t_1 = \frac{v-v_0}{a} = \frac{70 ~m/s}{3.2 ~m/s^2} = 22 ~s$
It takes 22 s until the rocket runs out of fuel.
(c) After the rocket runs out of fuel:
$\Delta y = \frac{v^2-v_0^2}{2a} = \frac{0 - (70 ~m/s)^2}{(2)(-9.8)} = 250 ~m$
The maximum altitude is 775 m + 250 m = 1025 m.
(d) After the rocket runs out of fuel:
$t_2 = \frac{v-v_0}{a} = \frac{0 - 70 ~m/s}{-9.8 ~m/s^2} = 7.1 ~s$
The total time to reach maximum altitude is $t_1 + t_2$ which is 22 s + 7.1 s = 29 s
(e) $v^2 = v_0^2 + 2ay$
$v = \sqrt{0 + (2)(9.8 ~m/s^2)(1025 ~m)} = 140 ~m/s$
The rocket strikes the earth with a velocity of 140 m/s.
(f) The time to drop from maximum altitude to the ground is:
$t_3 = \frac{v-v_0}{a} = \frac{140 ~m/s}{9.8 ~m/s^2} = 14 ~s$
The total time in the air is the time up to maximum altitude plus the time to drop back down, which is 29 s + 14 s = 43 s. The total time in the air is 43 s.
