A model rocket rises with constant acceleration to a height of 3.1 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 acceleration? C.) Find the height of the rocket 0.15 s after launch.
Express your answer using two significant figures. D.) Find the speed of the rocket 0.15 s after launch.
Hunter op
A.) To find the time it takes for the rocket to reach a height of 3.1 m, we can use the following kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (26.0 m/s)
- \( u \) is the initial velocity (0 m/s as the rocket started from rest)
- \( a \) is the acceleration
- \( s \) is the distance (3.1 m)
Rearranging the equation, we get:
\[ t = \frac{v-u}{a} \]
Substituting the values into the equation:
\[ t = \frac{26.0\, \text{m/s} - 0\, \text{m/s}}{a} \]
B.) To find the magnitude of the rocket's acceleration, we can use the same equation as above:
\[ a = \frac{v^2 - u^2}{2s} \]
Substituting the values into the equation:
\[ a = \frac{(26.0\, \text{m/s})^2 - (0\, \text{m/s})^2}{2(3.1\, \text{m})} \]
C.) To find the height of the rocket 0.15 s after launch, we can use the equation:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values into the equation:
\[ s = (0\, \text{m/s})(0.15\, \text{s}) + \frac{1}{2}a(0.15\, \text{s})^2 \]
D.) To find the speed of the rocket 0.15 s after launch, we can use the equation:
\[ v = u + at \]
Substituting the values into the equation:
\[ v = (0\, \text{m/s}) + a(0.15\, \text{s}) \]
(Note: To answer C and D, we need to know the value of the acceleration, which we haven't found yet. We'll find the acceleration in part B first and then substitute it into these equations.)
Now let's solve for the answers step-by-step:
A.) Time taken to reach a height of 3.1 m:
\[ t = \frac{26.0\, \text{m/s} - 0\, \text{m/s}}{a} \]
B.) Magnitude of the rocket's acceleration:
\[ a = \frac{(26.0\, \text{m/s})^2 - (0\, \text{m/s})^2}{2(3.1\, \text{m})} \]
C.) Height of the rocket 0.15 s after launch:
\[ s = (0\, \text{m/s})(0.15\, \text{s}) + \frac{1}{2}a(0.15\, \text{s})^2 \]
D.) Speed of the rocket 0.15 s after launch:
\[ v = (0\, \text{m/s}) + a(0.15\, \text{s}) \]
A.) To find the time it takes for the rocket to reach a height of 3.1 m, we can use the kinematic equation:
\[s = ut + \frac{1}{2}at^2 \]
where s is the displacement (height), u is the initial velocity, t is the time, and a is the acceleration.
Given:
s = 3.1 m
u = 0 m/s (initial velocity)
a = unknown (acceleration)
We need to solve for t, so we rearrange the equation to isolate t:
\[t = \sqrt{\frac{2s}{a}} \]
Using the information given, we can solve for t.
B.) To find the magnitude of the rocket's acceleration, we can use another kinematic equation:
\[v = u + at \]
Given:
v = 26.0 m/s (final velocity)
u = 0 m/s (initial velocity)
t = unknown (time)
a = unknown (acceleration)
We need to solve for a, so we rearrange the equation to isolate a:
\[a = \frac{v - u}{t} \]
Using the information given, we can solve for a.
C.) To find the height of the rocket 0.15 s after launch, we can use the equation derived in part A:
\[s = ut + \frac{1}{2}at^2 \]
where s is the displacement (height), u is the initial velocity, t is the time, and a is the acceleration.
Given:
u = 0 m/s (initial velocity)
t = 0.15 s (time)
a = unknown (acceleration)
We need to solve for s, so we plug in the known values and solve for s.
D.) To find the speed of the rocket 0.15 s after launch, we can use the equation derived in part B:
\[v = u + at \]
Given:
u = 0 m/s (initial velocity)
t = 0.15 s (time)
a = from part B (acceleration)
We need to solve for v, so we plug in the known values and solve for v.