A two-stage rocket is launched straight up from rest. It accelerates at 26.4 m/s/s for 3.92 seconds during the first stage. It accelerates at 12.3 m/s/s for 4.48 more seconds during the second stage. After second stage burnout the rocket comes under the influence of gravity only. Determine its height above launch point (in meters) after 21.2 seconds from launch.
after 1st stage:
final velocity __ Vf1 = 26.4 * 3.92
final height __ Hf1 = .5 * 26.4 * 3.92^2
after 2nd stage:
final velocity __ Vf2 = (12.3 * 4.48) + Vf1
final height __ Hf2 = (.5 * 12.3 * 4.48^2) + (Vf1 * 4.48) + Hf1
free-fall time is __ 21.2 s - 3.92 s - 4.48 s , or 12.8
H = (-.5 * g * 12.8^2) + (Vf2 * 12.8) + Hf2
To determine the height above the launch point after 21.2 seconds, we first need to find the velocities achieved during each stage.
During the first stage, the rocket accelerates at 26.4 m/s^2 for 3.92 seconds from rest. We can find the velocity attained using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and t is the time.
So, for the first stage, the velocity attained is:
v1 = 0 + (26.4 m/s^2) × (3.92 s) = 103.01 m/s.
Next, during the second stage, the rocket accelerates at 12.3 m/s^2 for an additional 4.48 seconds. The final velocity at the end of the second stage is:
v2 = v1 + (12.3 m/s^2) × (4.48 s) = 154.8816 m/s.
After the second stage burnout, the rocket comes under the influence of gravity only. To find the height reached, we need to calculate the distance traveled during the powered stages and then add the height gained under gravity.
During the first stage, the distance traveled can be calculated using the equation:
s1 = ut + (1/2)at^2,
where s1 is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
For the first stage:
s1 = (0 m/s) × (3.92 s) + (1/2) × (26.4 m/s^2) × (3.92 s)^2 = 204.2624 m.
During the second stage, the distance traveled can be calculated using the same equation:
s2 = v1t + (1/2)at^2,
where s2 is the distance traveled, v1 is the initial velocity at the start of the second stage, a is the acceleration during the second stage, and t is the time.
For the second stage:
s2 = (103.01 m/s) × (4.48 s) + (1/2) × (12.3 m/s^2) × (4.48 s)^2 = 640.8587 m.
Now, we need to calculate the distance traveled under the influence of gravity from the end of the second stage to 21.2 seconds. We will use the equation:
s3 = v2t + (1/2)gt^2,
where s3 is the distance traveled, v2 is the final velocity at the end of the second stage, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
For this phase:
s3 = (154.8816 m/s) × (21.2 s - 3.92 s - 4.48 s) + (1/2) × (-9.8 m/s^2) × (21.2 s - 3.92 s - 4.48 s)^2 = -2754.3408 m.
Finally, we can calculate the total height above the launch point by adding the distances traveled during each stage:
Total height = s1 + s2 + s3 = 204.2624 m + 640.8587 m - 2754.3408 m ≈ -1909.22 m.
However, the negative sign implies that the rocket is below the launch point after 21.2 seconds. This indicates that the rocket has fallen back to Earth and is in the negative direction (below the launch point).
Therefore, the height above the launch point after 21.2 seconds from launch is approximately -1909.22 meters.