A 2 stage rocket launches off at an initial angle of 75 degrees above the horizontal. The rocket has a mass of 250 kg, while the stage 1 has a mass of 125 kg, and the stage 2 booster has a mass of 50 kg. At ignition the stage 1 booster gives the rocket a thrust of 34,652 N for 14 seconds. Then the stage 1 booster falls off and the stage 2 booster immediately fires. The stage 2 booster gives the rocket a thrust of 15,000 N for a length of 8 seconds. Then the stage 2 booster falls off and the rocket is now a projectile for the rest of its flight until it crashes into the ground. Assume the rocket went in a straight lin during the boosters firing.
a) How high did the rocket go?
b) How dar out did the rocket go?
c) How long was it in the air?
To solve these problems, we need to break down the rocket's motion into different phases and use the principles of projectile motion. Let's go step by step:
a) To determine how high the rocket went, we need to find the maximum height reached during its flight. To do this, we'll use the kinematic equation for vertical motion:
y = y0 + v0y*t - 0.5*g*t^2
Where:
y is the vertical displacement or height
y0 is the initial vertical position (0 in this case)
v0y is the initial vertical velocity
t is the time
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Stage 1:
First, let's find the initial vertical velocity v0y for the stage 1 booster. We can use Newton's second law of motion:
F = m*a
Where:
F is the force (thrust) applied
m is the mass of the rocket + stage 1 booster
a = acceleration
Using the given values:
F = 34,652 N
m = 375 kg (250 kg rocket + 125 kg stage 1 booster)
We can rearrange the equation to find the acceleration:
a = F / m
a = 34,652 N / 375 kg
a = 92.27 m/s^2
Now, we can find the initial vertical velocity using the equation: v0y = a*t
v0y = 92.27 m/s^2 * 14 s
v0y = 1291.78 m/s
Now we can calculate the maximum height reached by the stage 1 booster using the kinematic equation:
y1 = y0 + v0y*t - 0.5*g*t^2
y1 = 0 + 1291.78 m/s * 14 s - 0.5 * 9.8 m/s^2 * (14 s)^2
y1 = 18,085.32 m - 960.4 m
y1 = 17,124.92 m
Therefore, the stage 1 booster reached a maximum height of 17,124.92 meters.
Stage 2:
Now, let's find the initial vertical velocity v0y for the stage 2 booster. Similarly, we'll use Newton's second law of motion:
F = m*a
Using the given values:
F = 15,000 N (thrust of stage 2 booster)
m = 300 kg (250 kg rocket + 50 kg stage 2 booster)
We can find the acceleration:
a = F / m
a = 15,000 N / 300 kg
a = 50 m/s^2
Now, we can find the initial vertical velocity using the equation: v0y = a*t
v0y = 50 m/s^2 * 8 s
v0y = 400 m/s
To find the maximum height for stage 2, we'll use the same kinematic equation:
y2 = y0 + v0y*t - 0.5*g*t^2
Since the rocket's initial vertical position for stage 2 is the height it reached in stage 1, y0 = 17,124.92 m:
y2 = 17,124.92 m + 400 m/s * 8 s - 0.5 * 9.8 m/s^2 * (8 s)^2
y2 = 17,124.92 m + 3200 m - 313.6 m
y2 = 20,011.32 m
Therefore, the stage 2 booster reached a maximum height of 20,011.32 meters.
b) To determine how far the rocket traveled horizontally, we need to find the horizontal displacement. Since the rocket goes in a straight line, the horizontal motion is uniform. We can use the equation:
x = v0x * t
where:
x is the horizontal displacement
v0x is the initial horizontal velocity
t is the total time in the air
Since the rocket is launched at an angle of 75 degrees above the horizontal, we need to find the horizontal component of the velocity. Using trigonometry:
v0x = v0 * cosθ
where:
v0 is the initial velocity of the rocket (unknown)
θ is the launch angle (75 degrees)
To find v0, we can use the equation of motion for vertical motion:
v0 = v0y + g*t
For stage 1:
v0 = 1291.78 m/s + 9.8 m/s^2 * 14 s
v0 = 1434.38 m/s
For stage 2:
v0 = 400 m/s + 9.8 m/s^2 * 8 s
v0 = 476.4 m/s
Now we can calculate the horizontal displacements:
For stage 1:
x1 = v0x * t = v0 * cosθ * t = 1434.38 m/s * cos(75°) * 14 s = 4,222.45 m
For stage 2:
x2 = v0x * t = v0 * cosθ * t = 476.4 m/s * cos(75°) * 8 s = 1,405.12 m
Total horizontal displacement:
x_total = x1 + x2 = 4,222.45 m + 1,405.12 m = 5,627.57 m
Therefore, the rocket traveled a distance of 5,627.57 meters horizontally.
c) The total time the rocket was in the air is the sum of the burn times for both stages. We add the time for stage 1 (14 s) and stage 2 (8 s):
Total time = 14 s + 8 s = 22 s
Therefore, the rocket was in the air for 22 seconds.