To solve this problem, we can use the kinematic equations of motion.
a) To find the maximum height attained by the rocket, we need to find the time it takes to reach that height. We can use the equation:
h = vi*t + (1/2)*a*t^2
where h is the maximum height attained, vi is the initial velocity (which is zero), a is the acceleration, and t is the time taken to reach the maximum height.
Plugging in the values given, we get:
26m = 0*t + (1/2)*12m/s^2*t^2
Simplifying the equation, we get:
t^2 = (2*26m) / 12m/s^2
t^2 = 3.5s^2
t = 1.87s
Now that we know the time taken to reach the maximum height, we can use another kinematic equation to find the maximum height:
v = vi + a*t
where v is the final velocity at the maximum height.
Plugging in the values given, we get:
v = 0 + 12m/s^2*1.87s
v ≈ 22.44m/s
Now we can find the maximum height using the equation:
h = vi*t + (1/2)*a*t^2
Plugging in the values given, we get:
h = 0*1.87s + (1/2)*12m/s^2*(1.87s)^2
h ≈ 26.2m
Therefore, the maximum height attained by the rocket is approximately 26.2 meters.
b) To find the speed of the rocket just before it hits the ground, we can use the equation:
v^2 = vi^2 + 2*a*h
where h is the maximum height attained, vi is the initial velocity (which is zero), a is the acceleration, and v is the final velocity just before hitting the ground.
Plugging in the values given, we get:
v^2 = 0 + 2*12m/s^2*26m
v^2 = 624m^2/s^2
v ≈ 25m/s
Therefore, the speed of the rocket just before it hits the ground is approximately 25 meters per second.
c) The total duration of the rocket's flight is the time taken to reach the maximum height plus the time taken to fall back