A model rocket launches with an initial velocity of 6 m/s, at an angle of 90 degrees (with respect to the horizon). The rocket's engine provides an acceleration of 2.2 m/s2, acting in the same direction as the initial velocity. After 20 seconds, the rocket engine cuts out.

What is the maximum height attained by the rocket?

at twenty seconds,

vf=vi+at=50.4 m/s check that.
h=(vi*20+1/2 a t^2)= 120+1.1*400=560 check that.

now, how much higher did it go, with an initial velocity of 50.4m/s

vf at top=0
vf^2=vi^2+2ad
0=50,4^2-19.8d solve for d

Now add d to 560m

To find the maximum height attained by the rocket, we need to calculate the vertical component of its motion separately. Here's how we can do it step by step:

1. Calculate the vertical velocity of the rocket at any given time using the equation:
Vy = Voy - gt

In this case, the initial vertical velocity (Voy) is equal to the initial velocity (6 m/s), as the rocket initially points straight up. The acceleration due to gravity (g) is approximately 9.8 m/s² and acts downward, so we take it as negative.

2. Find the time it takes for the rocket engine to cut out. In this case, the engine cuts out after 20 seconds.

3. Calculate the time it takes to reach the maximum height by setting Vy = 0 and solving for time. Rearranging the equation from step 1:
Voy - gt = 0
t = Voy / g

4. Calculate the maximum height above the ground using the equation:
h = Voy * t - (1/2) * g * t^2

Let's plug in the values and calculate the maximum height attained by the rocket:

Voy = 6 m/s
g = 9.8 m/s² (acceleration due to gravity)
t = 6 m/s / 9.8 m/s² = 0.612 seconds (time to reach maximum height)

h = (6 m/s) * (0.612 seconds) - (1/2) * (9.8 m/s²) * (0.612 seconds)^2

Evaluating the expression, we find:

h ≈ 1.845 meters

Therefore, the maximum height attained by the rocket is approximately 1.845 meters.