a model rocket rises with constant accerleration to a height of 3.2 m, at which point its speed is 26.0 m/s. (a) how much time does it take for the rocket to reach this height? (b) what was the magnitude of the rocket's acceration? (c) find the height and speed of the rocket 0.10 s after launch.
The equations that apply to rising bodies are
.....Vf = Vo - gt (the term “g” for acceleration due to gravity is assumed constant on, or near, the surface of the Earth)
.....d = Vo(t) - g(t^2)/2
.................2
.....Vf^2 = Vo^2 - 2gd
From Vf^2 = Vo^2 - 2gd
.....26^2 = 0^2 + 2(a-9.8)3.2 from which
.....a = 115.425m/s^2
From Vf = Vo - gt
.....26 = 0 + (a-9.8)t from which
t = .246sec.
At t = .10sec, from
Vf = Vo - gt
Vf = 0 + (a-9.8).10 From which
Vf = 10.56m/s
At t = .10sec, from
d = Vo(t) - g(t^2)/2 from which
d = o(.10) + (a-9.8)(.10)^2/2 from which
d = .528met.
To solve this problem, we'll use the kinematic equations for an object with constant acceleration. The four equations are:
1. v = u + at (equation for calculating final velocity)
2. s = ut + 0.5at^2 (equation for calculating displacement)
3. v^2 = u^2 + 2as (equation for calculating final velocity squared)
4. s = (u + v)t/2 (equation for calculating average displacement)
Given:
- Initial velocity (u) = 0 m/s (assuming the rocket starts from rest)
- Final velocity (v) = 26.0 m/s
- Height (s) = 3.2 m
- Time (t) = ? (for part a)
- Acceleration (a) = ? (for part b and c)
(a) How much time does it take for the rocket to reach this height?
We can use equation 2: s = ut + 0.5at^2, where u = 0 m/s.
Substituting the values, we get:
3.2 m = 0.5a(t^2)
To solve for t, we'll need the acceleration (a). Let's find it first.
(b) What was the magnitude of the rocket's acceleration?
We can use equations 1 and 3 together to find the acceleration.
From equation 3: v^2 = u^2 + 2as
Substituting the given values, we get:
(26.0 m/s)^2 = 0^2 + 2a(3.2 m)
Simplifying, we have:
676.0 m^2/s^2 = 6.4a
Dividing both sides by 6.4, we find:
a = 676.0 m^2/s^2 / 6.4
a ≈ 105.625 m/s^2
Now that we know the acceleration, we can solve for time (t).
Substituting the known values into equation 2:
3.2 m = 0.5(105.625 m/s^2)(t^2)
Rearranging the equation, we get:
0.5(105.625 m/s^2)(t^2) = 3.2 m
Dividing both sides by 0.5(105.625 m/s^2), we find:
t^2 = 3.2 m / (0.5(105.625 m/s^2))
Simplifying, we have:
t^2 ≈ 0.060633 m / (m/s^2)
t^2 ≈ 0.060633 s^2
Taking the square root of both sides, we find:
t ≈ √(0.060633 s^2)
t ≈ 0.2465 s (rounded to four decimal places)
Therefore, it takes approximately 0.2465 seconds for the rocket to reach a height of 3.2 meters.
(c) Find the height and speed of the rocket 0.10 seconds after launch.
To find the height and speed after 0.10 seconds, we can use equations 2 and 1, respectively.
Using equation 2 for height (displacement):
s = ut + 0.5at^2
Substituting the given values, we get:
s = 0 m/s * (0.10 s) + 0.5(105.625 m/s^2) * (0.10 s)^2
Simplifying, we have:
s ≈ 0.5253125 m (rounded to four decimal places)
Therefore, the height 0.10 seconds after launch is approximately 0.5253 meters.
Using equation 1 for speed (final velocity):
v = u + at
Substituting the values, we get:
v = 0 m/s + (105.625 m/s^2) * (0.10 s)
Simplifying, we have:
v ≈ 10.5625 m/s (rounded to four decimal places)
Therefore, the speed 0.10 seconds after launch is approximately 10.5625 m/s.
To solve these physics problems, we can use the equations of motion. The key equations we'll need are:
1. Position equation: s = ut + (1/2)at^2
2. Final velocity equation: v = u + at
3. Average velocity equation: v = (u + v)/2
where:
- s is the displacement (change in position)
- u is the initial velocity
- v is the final velocity
- a is the acceleration
- t is the time
Now let's address each part of the question.
(a) How much time does it take for the rocket to reach a height of 3.2m?
We need to find the time it takes for the rocket to reach a height of 3.2m. Given:
- Final velocity (v) = 26.0 m/s
- Initial velocity (u) = 0 (since the rocket starts from rest)
- Displacement (s) = 3.2m
Using the second equation (v = u + at), we have:
26.0 m/s = 0 + a*t
We can solve for time (t) using the first equation (s = ut + (1/2)at^2). Rearranging the equation and plugging in the given values:
3.2m = (1/2) a*t^2
6.4m = a*t^2
Now substitute the value of a*t from the second equation:
6.4m = (26.0 m/s) * t
Solving for t:
t = 6.4m / 26.0 m/s
t ≈ 0.246 seconds
So, it takes approximately 0.246 seconds for the rocket to reach a height of 3.2m.
(b) What was the magnitude of the rocket's acceleration?
Now that we have the time (t), we can use the second equation (v = u + at) with the given values to find the acceleration (a):
26.0 m/s = 0 + a * 0.246s
Solving for a:
a = (26.0 m/s) / 0.246s
a ≈ 105.69 m/s²
The magnitude of the rocket's acceleration is approximately 105.69 m/s².
(c) Find the height and speed of the rocket 0.10 seconds after launch.
Given:
- Time (t) = 0.10 seconds
- Initial velocity (u) = 0 (since the rocket starts from rest)
- Acceleration (a) = 105.69 m/s²
To find the height, we will use the first equation (s = ut + (1/2)at^2) by plugging in the given values:
s = (0 m/s) * 0.10s + (1/2) * 105.69 m/s² * (0.10s)^2
s = 0 + 0.5295 m
s = 0.53 m
Therefore, the height of the rocket 0.10 seconds after launch is approximately 0.53m.
To find the speed, we can use the second equation (v = u + at) by plugging in the given values:
v = (0 m/s) + 105.69 m/s² * 0.10s
v = 10.569 m/s
Therefore, the speed of the rocket 0.10 seconds after launch is approximately 10.569 m/s.