A small rocket is launched vertically, attaining a maximum speed at burnout of 1.0 102 m/s and thereafter coasting straight up to a maximum altitude of 1482 m. Assuming the rocket accelerated uniformly while the engine was on, how long did it fire and how high was it at engine cutoff? Ignore air friction.
Someone will glady critique your work.
That's why I posted here in the first place. I don't know how to even start it.
Well, it sounds like this rocket really went for the sky! But let me calculate the answers for you, using my impressive math skills... *puts on clown wig*
First, let's find out how long the rocket fired. To do that, we need to determine the time it took for the rocket to reach its maximum speed. We'll use the equation:
v = u + at
Where 'v' is the final velocity (1.0 x 10^2 m/s), 'u' is the initial velocity (0 m/s since it started at rest), 'a' is the acceleration, and 't' is the firing time.
Since the rocket accelerated uniformly, the acceleration during this phase is the same throughout. So we can rewrite the equation as:
1.0 x 10^2 = 0 + a * t
Now, we need to find 'a'. We know that acceleration is a change in velocity divided by time, but in this case, we have the final velocity and need to find acceleration. So we rearrange the equation:
a = (1.0 x 10^2) / t
Next, we'll use the equation for displacement during uniform acceleration:
s = ut + (1/2) a t^2
Since 's' is the maximum altitude (1482 m), 'u' is 0 m/s, and 't' is the firing time, we can write the equation as:
1482 = 0 + (1/2) * a * t^2
Let's plug in the value of 'a' that we found earlier:
1482 = 0 + (1/2) * (1.0 x 10^2) / t * t^2
Now, let's simplify that expression:
1482 = (1/2) * t * t
2964 = t^2
t = sqrt(2964)
t ≈ 54.4 seconds
Therefore, the rocket fired for approximately 54.4 seconds. Now let's find the height at engine cutoff, which is simply the maximum altitude it reached:
Height at engine cutoff = 1482 meters
So, the rocket fired for about 54.4 seconds and reached an altitude of 1482 meters before the engine cut off. I hope my calculations didn't go over your head!
To find the time the rocket was fired and the height at engine cutoff, we can use the equations of motion.
1. First, let's find the time the rocket was fired (the burnout time). We know that the rocket attained a maximum speed of 1.0 * 10^2 m/s.
To find the burnout time, we can use the equation:
v = u + at
Where:
v = final velocity = 1.0 * 10^2 m/s (maximum speed at burnout)
u = initial velocity = 0 m/s (rocket starts from rest)
a = acceleration = unknown (assumed to be uniform)
t = time taken to reach burnout
Rearranging the equation, we get:
t = (v - u) / a
Since the rocket starts from rest (u = 0), the equation simplifies to:
t = v / a
2. Next, let's find the height at engine cutoff. We know that the rocket reaches a maximum altitude of 1482 m.
The equation for calculating the maximum height (H) is:
H = ut + (1/2) * a * t^2
Where:
H = maximum height = 1482 m
u = initial velocity = 0 m/s (rocket starts from rest)
a = acceleration = unknown (assumed to be uniform)
t = time taken to reach maximum height (burnout time)
Since the rocket starts from rest (u = 0), the equation simplifies to:
H = (1/2) * a * t^2
Now we have two equations with two unknowns (a and t). We can solve them simultaneously to find the values.
1. Substitute the values of v and u into the equation for t:
t = (1.0 * 10^2 m/s) / a
2. Substitute the value of H into the equation for H:
1482 m = (1/2) * a * (1.0 * 10^2 m/s / a)^2
Simplifying the equation:
1482 m = (1/2) * a * (1.0 * 10^2 m/s)^2 / a^2
Cross-multiplying and rearranging:
2964 m * a^2 = 5000 m^2/s^2
Dividing both sides by 2964 m:
a^2 = 5000 m^2/s^2 / 2964 m
Taking the square root of both sides:
a = sqrt(5000 m^2/s^2 / 2964 m) ≈ 3.18 m/s^2
Now that we have the value of a, we can substitute it back into the equation for t to find the burnout time.
t = (1.0 * 10^2 m/s) / (3.18 m/s^2)
t ≈ 31.45 seconds
So, the rocket was fired for approximately 31.45 seconds and reached a height of 1482 meters.