A model rocket is launched straight upward
with an initial speed of 51.6 m/s. It accelerates with a constant upward acceleration of
1.96 m/s2 until its engines stop at an altitude
of 230 m.
What is the maximum height reached by
the rocket? The acceleration of gravity is
9.81 m/s2.
Answer in units of m
Well, let's see... the rocket starts with an initial velocity of 51.6 m/s and accelerates upwards at 1.96 m/s² until it reaches an altitude of 230 m.
To find the maximum height reached by the rocket, we can use the kinematic equation:
vf² = vi² + 2ad
Where:
vf = final velocity (which is 0 m/s since the engines stop)
vi = initial velocity (51.6 m/s)
a = acceleration (1.96 m/s²)
d = displacement (maximum height reached)
Plugging in the values, we get:
0² = (51.6)² + 2(1.96)d
Simplifying that, we have:
0 = 2656.56 + 3.92d
Now, let's solve for d:
3.92d = -2656.56
d ≈ -677.14 m
Oops! It seems like there's been an error. The negative value doesn't make sense for the maximum height. I apologize for my miscalculation. Let's try this again:
vf² = vi² + 2ad
0² = (51.6)² + 2(-9.81)d
0 = 2664.96 - 19.62d
19.62d = 2664.96
d ≈ 136 m
There we go! The maximum height reached by the rocket is approximately 136 m. Phew, I'm glad we avoided that pesky negative answer. Keep aiming high, rocket! It's just a shame you couldn't reach the moon for some cheese.
To find the maximum height reached by the rocket, we can use the equations of motion. Let's break down the steps:
Step 1: Find the time taken for the rocket to reach its maximum height.
Using the equation of motion:
vf = vi + at,
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, the final velocity will be 0 m/s since the rocket comes to a stop. The initial velocity (vi) is 51.6 m/s, and the acceleration (a) is -9.81 m/s^2 (negative because the acceleration is in the opposite direction of the motion).
Plugging in the given values:
0 = 51.6 - 9.81t
Solving for t:
9.81t = 51.6
t = 51.6 / 9.81
t ≈ 5.26 s
Step 2: Calculate the maximum height using the equation of motion:
d = vit + (1/2)at^2,
where d is the displacement, vi is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, the displacement will be the maximum height itself. The initial velocity (vi) is 51.6 m/s, the acceleration (a) is -9.81 m/s^2, and the time (t) is 5.26 s (calculated in the previous step).
Plugging in the given values:
d = (51.6)(5.26) + (1/2)(-9.81)(5.26)^2
Simplifying:
d ≈ 271.14 - 133.55
d ≈ 137.59 m
Therefore, the maximum height reached by the rocket is approximately 137.59 meters.
To find the maximum height reached by the rocket, we can use the equations of motion. Let's break down the problem into different stages:
Stage 1: Acceleration phase
During this phase, the rocket is accelerating upward at a constant rate of 1.96 m/s². We need to find the time it takes for the rocket to reach its maximum height during this stage.
The equation to find the time (t) it takes for an object to reach a certain height (h) under constant acceleration is:
h = ut + 0.5at²
Where:
- h is the height
- u is the initial velocity (51.6 m/s)
- a is the acceleration (1.96 m/s²)
- t is the time
We want to find t when h is at its maximum. At maximum height, the final velocity will be 0 m/s.
0 = 51.6 + (1.96)t
Rearranging the equation, we get:
t = -51.6 / 1.96
Solving this equation, we find t ≈ -26.33 seconds. However, time cannot be negative, so we discard the negative value. Therefore, the time it takes to reach maximum height during the acceleration phase is approximately 26.33 seconds.
Stage 2: Deceleration phase
After the engines stop at an altitude of 230 m, the rocket continues its motion upward with an initial velocity of 0 m/s due to the deceleration caused by gravity. We need to find the additional height gained during this phase.
The equation to find the additional height (h) gained under constant acceleration due to gravity (g) is:
h = ut + 0.5gt²
Where:
- h is the height gained
- u is the initial velocity (0 m/s)
- g is the acceleration due to gravity (9.81 m/s²)
- t is the time
Since the rocket is starting from rest, the initial velocity (u) is 0. The equation simplifies to:
h = 0.5gt²
Plugging in the values:
h = 0.5 * 9.81 * (26.33)²
Simplifying the equation, we find:
h ≈ 3384.9 meters
Therefore, the additional height gained during the deceleration phase is approximately 3384.9 meters.
Stage 3: Total maximum height
To find the maximum height reached by the rocket, we add the heights gained during both stages:
Maximum height = height gained during acceleration phase + height gained during deceleration phase
Maximum height = 230 + 3384.9
Maximum height ≈ 3614.9 meters
Thus, the maximum height reached by the rocket is approximately 3614.9 meters.