Answer:
a) The velocity of the rocket when it runs out of fuel is 67 m/s
b) It takes the rocket 21 s to reach an altitude of 725 m
c) The maximum altitude of the rocket is 954 m
d) and e) It takes the rocket 28 s to reach the maximum altitude.
f) The rocket is 42 s in the air.
Explanation:
The equations for the height of the rocket are the following:
y = y0 + v0 · t + 1/2 · a · t²
and, when the rocket runs out of fuel:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the rocket at time t
y0 = initial height
v0 = initial velocity
a = acceleration due to engine of the rocket.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
t = time
Let´s place the origin of the frame of reference at the launching point.
b) First, let´s calculate the time it takes the rocket to reach an altitude of 725 m:
y = y0 + v0 · t + 1/2 · a · t² (y0 = 0, v0 = 0)
725 m = 1/2 · 3.2 m/s² · t²
725 m / 1/2 · 3.2 m/s² = t²
<u>t = 21 s</u>
a)The velocity of the rocket is given by the following equation:
v = v0 + a · t (v0 = 0)
v = 3.2 m/s² · 21 s = <u>67 m/s</u>
c) After the rocket runs out of fuel, it starts slowing down until it comes to stop and starts to fall. Let´s calculate the time it takes the rocket to stop after running out of fuel. At that time, the rocket will be at its maximum height.
v = v0 + g · t
0 m/s = 67 m/s - 9.8 m/s² · t
-67 m/s / -9.8 m/s² = t
t = 6.8 s
The altitude reached after running out of fuel will be:
y = y0 + v0 · t + 1/2 · g · t²
y = 725 m + 67 m/s · 6.8 s - 1/2 · 9.8 m/s² · (6.8 s)²
y = 954 m
d) and e)
The maximum altitude will be reached after (21 s + 6.8 s) 28 s
f) Now, let´s calculate how much time it takes the rocket to return to the ground (y = 0) from the max-altitude:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 954 m + 0 m/s · t - 1/2 · 9.8 m/s² ·t²
-954/- 1/2 · 9.8 m/s² = t²
t = 14 s
The rocket was (28 s + 14 s) 42 s in the air.
