A model rocket leaves the ground, heading straight up at 45m/s .
Part A
What's its maximum altitude?
Express your answer to two significant figures and include the appropriate units.
Part B
Find its velocity at 1 s?
Express your answer to two significant figures and include the appropriate units.
Part C
Find its altitude at 1 s?
Express your answer to two significant figures and include the appropriate units.
Part D
Find its velocity at 4 s?
Express your answer to two significant figures and include the appropriate units.
Part E
Find its altitude at 4 s?
Express your answer to two significant figures and include the appropriate units.
Part F
Find its velocity at 7 s?
Express your answer to two significant figures and include the appropriate units.
Part G
Find its altitude at 7 s?
Express your answer to two significant figures and include the appropriate units.
well, you know that the height is
y = 45t - 4.9t^2
Now just use that parabola to answer the questions.
To solve this problem, we'll use the equations of motion for constant acceleration:
1. Displacement (s) = Initial velocity (v0) * Time (t) + 1/2 * Acceleration (a) * Time (t)^2
2. Final velocity (v) = Initial velocity (v0) + Acceleration (a) * Time (t)
Given information:
Initial velocity (v0) = 45 m/s
Acceleration (a) = -9.8 m/s^2 (due to gravity)
Part A: Maximum altitude
To find the maximum altitude, we need to find the time it takes for the rocket to reach its peak and then use that time in the equation for displacement. At the maximum altitude, the rocket's final velocity is 0 m/s.
Using equation (2):
Final velocity (v) = v0 + a * t
0 = 45 + (-9.8) * t
9.8t = 45
t = 45 / 9.8
Using equation (1):
Displacement (s) = v0 * t + 1/2 * a * t^2
s = 45 * (45 / 9.8) + 1/2 * (-9.8) * (45 / 9.8)^2
Part B: Velocity at 1 s
Using equation (2):
v = v0 + a * t
Part C: Altitude at 1 s
Using equation (1):
s = v0 * t + 1/2 * a * t^2
Part D: Velocity at 4 s
Using equation (2):
v = v0 + a * t
Part E: Altitude at 4 s
Using equation (1):
s = v0 * t + 1/2 * a * t^2
Part F: Velocity at 7 s
Using equation (2):
v = v0 + a * t
Part G: Altitude at 7 s
Using equation (1):
s = v0 * t + 1/2 * a * t^2
Now, let's calculate the values.
To solve these problems, we can use the equations of motion that relate the velocity, time, and acceleration of an object. In this case, the rocket is moving straight up and we are given its initial velocity. We can assume that the acceleration is due to gravity, which is approximately 9.8 m/s^2.
Part A
The maximum altitude of the rocket can be found by using the formula for displacement (height) of an object under constant acceleration: Δy = v0t + (1/2)at^2, where Δy is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration. Plugging in the values:
Δy = (45 m/s)(t) + (1/2)(-9.8 m/s^2)(t^2)
To find the maximum altitude, we need to find the time it takes for the rocket to reach its peak. At the highest point, the rocket's vertical velocity will be zero. So, we can set the equation for velocity equal to zero and solve for t:
0 = 45 m/s - 9.8 m/s^2(t)
Solving this equation gives t = 4.6 s. Now we can plug this value into the equation for displacement:
Δy = (45 m/s)(4.6 s) + (1/2)(-9.8 m/s^2)(4.6 s)^2
Δy ≈ 104 m
Therefore, the maximum altitude of the rocket is approximately 104 meters.
Part B
To find the velocity of the rocket at 1 second, we can plug the given time into the equation for velocity: v = v0 + at.
v = 45 m/s + (-9.8 m/s^2)(1 s)
v ≈ 35 m/s
Therefore, the velocity of the rocket at 1 second is approximately 35 meters per second.
Part C
The altitude of the rocket at 1 second can be found by using the equation for displacement mentioned earlier (Δy = v0t + (1/2)at^2):
Δy = (45 m/s)(1 s) + (1/2)(-9.8 m/s^2)(1 s)^2
Δy ≈ 38 m
Therefore, the altitude of the rocket at 1 second is approximately 38 meters.
Part D
To find the velocity of the rocket at 4 seconds, we can plug the given time into the equation: v = v0 + at.
v = 45 m/s + (-9.8 m/s^2)(4 s)
v ≈ 4 m/s
Therefore, the velocity of the rocket at 4 seconds is approximately 4 meters per second.
Part E
The altitude of the rocket at 4 seconds can be found using the equation for displacement: Δy = v0t + (1/2)at^2.
Δy = (45 m/s)(4 s) + (1/2)(-9.8 m/s^2)(4 s)^2
Δy ≈ 66 m
Therefore, the altitude of the rocket at 4 seconds is approximately 66 meters.
Part F
To find the velocity of the rocket at 7 seconds, we can again use the equation: v = v0 + at.
v = 45 m/s + (-9.8 m/s^2)(7 s)
v ≈ -24 m/s
Therefore, the velocity of the rocket at 7 seconds is approximately -24 meters per second, indicating that the rocket is moving downward.
Part G
Since the rocket is moving downward at 7 seconds, its altitude will be negative. To find the magnitude of the altitude, we can use the equation for displacement: Δy = v0t + (1/2)at^2.
Δy = (45 m/s)(7 s) + (1/2)(-9.8 m/s^2)(7 s)^2
Δy ≈ -132 m
Therefore, the altitude of the rocket at 7 seconds is approximately -132 meters, indicating that it is 132 meters below the ground.