A model rocket is launched straight upward with an initial speed of 60.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 130 m.
(a) What is the maximum height reached by the rocket?
m
(b) How long after lift-off does the rocket reach its maximum height?
s
(c) How long is the rocket in the air?
s
I would use energy concepts.
Launch energy: 1/2 m 60^2
engine energy: F*d=ma*d=m*2*130
work done on gravity at the top: mg*h
mg*h=1/2 m 60^2+m*260 solve for h.
For b) find time to the rocket burns out, then add the glide time to the final heights.
c. Add the fall time.
could you possibly explain A. im still not understanding it.
I solved for h and got 210.2. What am I doing wrong?
To find the answers to these questions, we can use the equations of motion for uniformly accelerated linear motion. Let's break it down step by step:
First, let's find the time it takes for the rocket to reach its maximum height (b).
We can use the following equation:
v = u + at
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (60.0 m/s)
a = acceleration (-2.00 m/s^2, since it's moving in the opposite direction of the positive y-axis)
t = time
Rearranging the equation to solve for time:
t = (v - u) / a
Substituting the given values, we get:
t = (0 - 60.0) / -2.00
Simplifying, we find:
t = 30.0 seconds
So, it takes 30.0 seconds for the rocket to reach its maximum height.
Next, let's find the maximum height reached by the rocket (a).
We can use the following equation:
s = ut + (1/2)at^2
Where:
s = displacement (unknown, maximum height)
u = initial velocity (60.0 m/s)
a = acceleration (-2.00 m/s^2)
t = time (30.0 seconds)
Substituting the given values, we get:
s = (60.0 * 30.0) + (1/2) * (-2.00) * (30.0)^2
Simplifying, we find:
s = 900 - 900.0
Therefore, the maximum height reached by the rocket is 0 m.
Lastly, let's find the total time the rocket is in the air (c).
Since the rocket starts at the ground and ends at 130 m, we can consider the up and down journey as one complete trip. Therefore, we can double the time it took for the rocket to reach its maximum height.
So, the total time the rocket is in the air is:
2 * 30.0 = 60.0 seconds
Therefore, the rocket is in the air for 60.0 seconds.