A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.20 m/s2 until it reaches an altitude of 960 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

(a) How long is the rocket in motion above the ground?

(b) What is its maximum altitude?

(c) What is its velocity just before it collides with the Earth?

I will be happy to critique your work.

A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?

To find the answers to the given questions, we will break down the motion of the rocket into two separate phases: the phase when the engines are operational and the phase when the rocket is in free fall.

Let's begin with the first phase when the engines are operational:

(a) How long is the rocket in motion above the ground?
To find the time during the engines' operation, we need to determine the time it takes for the rocket to reach an altitude of 960 m.
We can use the equation of motion:

s = ut + (1/2)at^2

where:
s = displacement (960 m)
u = initial velocity (80.2 m/s)
a = acceleration (4.20 m/s^2)
t = time

Rearranging the equation, we get:

960 = 80.2t + (1/2)(4.20)(t^2)

Rearranging and simplifying, we have:

2.10t^2 + 80.2t - 960 = 0

Solving this quadratic equation, we find two possible values for t. However, we are only interested in the positive solution since time cannot be negative.

Using the quadratic formula, we get:

t = (-80.2 ± sqrt((80.2)^2 - 4(2.10)(-960)) / (2.10)

t ≈ 8.45 s

Therefore, the rocket is in motion for approximately 8.45 seconds during the engines' operation.

(b) What is its maximum altitude?
To determine the maximum altitude, we need to calculate how high the rocket reaches during the engines' operation.
We can use the equation of motion:

v^2 = u^2 + 2as

where:
v = final velocity (unknown)
u = initial velocity (80.2 m/s)
a = acceleration (4.20 m/s^2)
s = displacement (unknown)

We need to solve for s, which represents the maximum altitude.

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Since the rocket's engines fail at the maximum altitude, its final velocity before free fall is 0 m/s.

Plugging in the known values, we have:

s = (0^2 - 80.2^2) / (2(-9.80))

Simplifying, we get:

s ≈ 5,247.96 m

Therefore, the rocket reaches a maximum altitude of approximately 5,247.96 meters.

Now, let's move on to the second phase when the rocket is in free fall:

(c) What is its velocity just before it collides with the Earth?
In this phase, the rocket is in free fall with an acceleration of -9.80 m/s^2.

To find the velocity just before collision, we can use the equation of motion:

v^2 = u^2 + 2as

where:
v = final velocity (unknown)
u = initial velocity (0 m/s)
a = acceleration (-9.80 m/s^2)
s = displacement (960 m)

We need to solve for v.

Plugging in the known values, we have:

v^2 = 0^2 + 2(-9.80)(960)

Simplifying, we get:

v^2 = -18,816

Since velocity cannot be negative in this context, we discard the negative value.

Taking the square root of the positive value, we get:

v ≈ 137.28 m/s

Therefore, the velocity just before the rocket collides with the Earth is approximately 137.28 m/s.

To solve these questions, we need to break down the motion of the rocket into two separate phases: the first phase when the engines are operating, and the second phase when the rocket is in free fall. Let's calculate the answers step by step:

(a) How long is the rocket in motion above the ground?

In the first phase, the rocket is accelerating upward, so we can use the kinematic equation: v = u + at, where:
v = final velocity (which we don't know)
u = initial velocity = 80.2 m/s
a = acceleration = 4.20 m/s²
t = time (which we want to find)

Since the rocket starts from rest (launch from the ground), we can substitute these values into the equation and solve for t:

v = u + at
0 + (4.2 * t) = 80.2
4.2t = 80.2
t = 80.2 / 4.2
t ≈ 19.10 seconds

So, the rocket is in motion above the ground for approximately 19.10 seconds.

(b) What is its maximum altitude?

To find the maximum altitude, we need to calculate the displacement during the first phase of motion. We can use the equation: s = ut + (1/2)at², where:
s = displacement (which we want to find)
u = initial velocity = 80.2 m/s
a = acceleration = 4.20 m/s²
t = time calculated in part (a) = 19.10 seconds

Substituting the values into the equation:

s = (80.2 * 19.1) + (0.5 * 4.2 * (19.1)²
s ≈ 1534.92 meters

So, the rocket's maximum altitude is approximately 1534.92 meters.

(c) What is its velocity just before it collides with the Earth?

In the second phase, when the engines fail, the rocket goes into free fall. The acceleration in free fall is -9.80 m/s² (negative due to gravity pulling downward). We can use the equation v = u + at, where:
v = final velocity (which we want to find)
u = initial velocity (which we know from the end of the first phase)
a = acceleration in free fall = -9.80 m/s²
t = time (which we want to find)

Since the initial velocity is the final velocity of the first phase (it's continuous), we substitute the already known values and solve for t:

v = 80.2 + (-9.80 * t)

When the rocket collides with the Earth, its final velocity (v) will be zero. So:

0 = 80.2 - 9.80t
9.80t = 80.2
t = 80.2 / 9.80
t ≈ 8.18 seconds

Now we can calculate the final velocity:

v = 80.2 + (-9.80 * 8.18)
v ≈ -0.34 m/s

So, the velocity just before the rocket collides with the Earth is approximately -0.34 m/s. The negative sign indicates that the rocket is moving downward.