You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 17 m/s2. After 20 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 12 km above the ground.

(a) What is the highest point your rocket reaches?


(b) Determine the total time the rocket is in the air.


(c) Find the speed of the rocket just before it hits the ground.

Thrust phase:

v = a t
v = 17 t
v at 20 s = 17 * 20 = 340 m/s (Vi for next phase)
h = (1/2) a t^2 = (1/2)(17)(400) = 3,400 m

coast up phase:
v = Vi - a t
v = 0 at top
0 = 340 - 9.81 t
t = 34.7 seconds coasting up
now 54.7 seconds aloft
how high?
h = 3,400 + Vi t - 4.9 t^2
h = 3,400 +340(34.7) -4.9(34.7)^2
= 3,400 + 11,784 - 5,886
= 9,298 m made like 9 km, not 12

now how long to fall
0 = 9.298 - 4.9 t^2
t = 43.6 s
so 54.7 + 43.6 = 98.3 seconds total aloft

(c) v = sqrt (2 g h)

To solve this problem, we can break it down into different parts. Let's start by finding the highest point the rocket reaches.

(a) What is the highest point your rocket reaches?

To find the highest point, we need to determine the time it takes for the rocket to reach its maximum height. We know that the rocket is accelerating with a constant upward acceleration of 17 m/s².

Using the formula for displacement, we have:

s = ut + (1/2)at²

Where:
s = displacement (change in height)
u = initial velocity (0 m/s)
t = time taken
a = acceleration (17 m/s²)

Since the rocket starts with an initial velocity of 0 m/s, the equation simplifies to:

s = (1/2)at²

We want to find the time it takes for the rocket to reach its highest point. At this point, the final velocity will be 0 m/s. Thus, we can set the final velocity equal to 0 and solve for time:

0 = u + at

0 = 0 + 17t

17t = 0

t = 0

This tells us that it takes 0 seconds for the rocket to reach its highest point. However, this doesn't seem correct. In order to reach a maximum height, the rocket needs some time to accelerate.

Upon closer inspection, we realize that the rocket is accelerating upwards until the engine shuts off at 20 seconds. After the engine shuts off, the rocket will continue rising momentarily but eventually starts falling due to gravity. Therefore, the highest point is reached at the moment the engine shuts off.

Therefore, the highest point the rocket reaches is at 20 seconds. To find the height, we can substitute this time into our formula:

s = (1/2)at²

s = (1/2)(17 m/s²)(20 s)²

s = (1/2)(17 m/s²)(400 s²)

s = 3400 m

The highest point the rocket reaches is 3400 meters above the ground.

(b) Determine the total time the rocket is in the air.

To determine the total time the rocket is in the air, we need to consider the time it takes for the rocket to reach its highest point, as well as the time it takes for the rocket to fall back to the ground.

The time spent going up to the highest point is 20 seconds, as we established earlier.

To find the time it takes for the rocket to fall back to the ground, we can use the equation for displacement:

s = ut + (1/2)at²

Since the rocket is falling downwards, the acceleration due to gravity will be negative (-9.8 m/s²). We need to find the time it takes for the displacement to be -12,000 meters (12 km).

-12,000 = 0 + (1/2)(-9.8 m/s²)t²

-24,000 = -4.9t²

t² = 24,000 / 4.9

t² = 4897.96

t ≈ √(4897.96)

t ≈ 69.99 s (rounded to the nearest hundredth)

Therefore, the total time the rocket is in the air is approximately 20 seconds (upward) + 69.99 seconds (falling), which is approximately 89.99 seconds (rounded to the nearest hundredth).

(c) Find the speed of the rocket just before it hits the ground.

To find the speed of the rocket just before it hits the ground, we can use the equation for final velocity:

v = u + at

In this case, the initial velocity is 0 m/s (since the rocket falls from rest), the acceleration is -9.8 m/s², and the time is the same as the total time in part (b), which is approximately 89.99 seconds.

v = 0 + (-9.8 m/s²)(89.99 s)

v ≈ -881.901 m/s (rounded to the nearest thousandth)

Note that the negative sign indicates the velocity is directed downwards since the rocket is falling.

Therefore, the speed of the rocket just before it hits the ground is approximately 881.901 m/s (rounded to the nearest thousandth).