A model rocket is launched straight upward with an initial speed of 47.8 m/s. It accelerates with a constant upward acceleration of 1.83 m/s2 until its engines stop at an altitude of 163 m. What is the maximum height reached by the rocket?

Vf^2 = Vo^2 + 2ad,

Vf^2 = (47.8)^2 + 2*1.83*163 = 2881.42,
Vf = 53.7m/s.

h = 163 + (Vf^2 - Vo^2) / 2g,
h = 163 + (0 - (53.7)^2) / -19.6 = 310m.

To find the maximum height reached by the rocket, we need to determine the time it takes for the engines to stop, and then use that time to calculate the rocket's height at that point.

Step 1: Calculate the time it takes for the rocket's engines to stop.
We can use the following kinematic equation to find the time it takes for the rocket to reach the altitude of 163 m:

h = h₀ + v₀t + (1/2)at²

Where:
h = final height (163 m)
h₀ = initial height (0 m)
v₀ = initial velocity (47.8 m/s)
t = time
a = acceleration (1.83 m/s²)

Rearranging the equation, we have:
163 = 0 + 47.8t + (1/2)(1.83)t²

Step 2: Solve for time (t).
By rearranging and simplifying the equation, we get a quadratic equation in terms of t:

0.915t² + 47.8t - 163 = 0

We can solve this quadratic equation using the quadratic formula or factoring. Let's use the quadratic formula:

t = (-b ± sqrt(b² - 4ac)) / 2a

In this case:
a = 0.915
b = 47.8
c = -163

Substituting these values into the quadratic formula, we can calculate the time (t).

Step 3: Calculate the maximum height reached by the rocket.
With the value of t, we can now find the maximum height reached by the rocket. We can use the kinematic equation again, this time with a negative acceleration (-1.83 m/s²) to represent the rocket's deceleration after the engines stop.

Using the equation:
h = h₀ + v₀t + (1/2)at²

Where:
h = maximum height
h₀ = initial height (0 m)
v₀ = initial velocity (47.8 m/s)
t = time
a = deceleration (-1.83 m/s²)

Substituting the known values, we can calculate the maximum height reached by the rocket.

After performing these calculations, we will find the maximum height reached by the rocket.