A model rocket is launched straight upward with an initial speed of 45.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 160 m.
A. what is the max height reached by the rocket?
B. How long after liftoff does the rocket reach its maximum height?
c. How long is the rocket in the air?
stage one:
h = 45 t + (1/2)(1.5) t^2 = 160
solve for t
.75 t^2 + 45 t - 160 = 0
t = 3.367 seconds
v = 45 + 1.5 t = 45 + 1.5(3.367) = 50.05 m/s
stage 2
v = Vi - g t
at top v = 0
9.81 t = 50.05
t = 5.1 seconds drifting up
h = 160 + 50.05(5.1)-4.9(5.1)^2
= 160 + 255 - 127
= 287 meters at the top
time so far = 3.4+5.1 = 8.5 seconds
stage 3, ignominious fall to ground
h = 287
so
287 = 4.9 t^2
t = 7.65 second fall
so
time in air = 8.5 + 7.7 = 16.2 seconds
To solve this problem, we can use the equations of motion. We know the initial velocity (u), the final velocity (v), the acceleration (a), and the displacement (s).
Given:
Initial velocity (u) = 45.0 m/s
Acceleration (a) = 1.50 m/s^2
Final displacement (s) = 160 m
A. To find the maximum height reached by the rocket, we need to find the displacement when the velocity becomes zero.
Using the equation v^2 = u^2 + 2as:
0^2 = (45.0 m/s)^2 + 2(1.50 m/s^2)(s)
0 = (2025 m^2/s^2) + 3s
3s = -2025 m^2/s^2
s = (-2025 m^2/s^2) / 3
s = -675 m^2/s^2
We can ignore the negative sign since it represents the downward direction. Thus, the maximum height reached by the rocket is 675 meters.
B. To find the time it takes for the rocket to reach its maximum height, we need to use the equation v = u + at and solve for time (t).
At the maximum height, the velocity becomes zero (v = 0).
0 = 45.0 m/s + (1.50 m/s^2)(t)
-45.0 m/s = 1.50 m/s^2
t = (-45.0 m/s) / (1.50 m/s^2)
t = -30 s^2/m
Again, we can ignore the negative sign. So, it takes 30 seconds for the rocket to reach its maximum height.
C. To find the total time the rocket is in the air, we can double the time it takes to reach the maximum height since the rocket will come back down.
Total time = 2(t)
Total time = 2(30 s)
Total time = 60 seconds
Therefore, the rocket is in the air for 60 seconds.
To find the maximum height reached by the rocket, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
where:
- vf is the final velocity (which is 0 m/s when the rocket reaches its maximum height)
- vi is the initial velocity (45.0 m/s)
- a is the acceleration (1.50 m/s^2)
- d is the displacement (the maximum height, which we need to find)
Rearranging the equation, we have:
0 = (45.0 m/s)^2 + 2(1.50 m/s^2)d
Simplifying and solving for d:
0 = 2025 m^2/s^2 + 3d
-2025 m^2/s^2 = 3d
d = -2025 m^2/s^2 / 3
Since we are dealing with a vertical displacement, the value of d will be positive. Therefore, we can take the absolute value of d:
d = |2025 m^2/s^2 / 3|
Now we can calculate the maximum height reached by the rocket:
A. The max height reached by the rocket is d = |2025 m^2/s^2 / 3| = 675 m. (Note: We take the absolute value to ensure a positive height.)
To find the time it takes for the rocket to reach its maximum height, we can use the equation:
vf = vi + at
where:
- vf is the final velocity (0 m/s)
- vi is the initial velocity (45.0 m/s)
- a is the acceleration (1.50 m/s^2)
- t is the time (which we need to find)
Rearranging the equation, we have:
0 = 45.0 m/s + (1.50 m/s^2)t
Solving for t:
-45.0 m/s = (1.50 m/s^2)t
t = -45.0 m/s / 1.50 m/s^2
t = -30 s
Here, we have a negative value for time, which indicates that we need to consider the time as the magnitude only:
B. The rocket reaches its maximum height 30 seconds after liftoff. (Note: We disregard the negative sign and consider the magnitude of time.)
Finally, to find how long the rocket is in the air, we need to consider the time it takes to reach the maximum height and then return to the ground. Since the rocket decelerates with the same 1.50 m/s^2 (in the opposite direction) when it starts coming down, the time to reach the ground will be twice the time to reach the maximum height.
C. The rocket is in the air for 2 * 30 s = 60 seconds.