During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.45 m/s2 . When it is 250 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).
1) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? Answer in meters (m).
2) What total distance did the canister travel between its release and its crash onto the launch pad? Answer in meters (m).
how long does it take to rise to 250m?
s = 1/2 at^2, so
3.45/2 t^2 = 250
t = 12 seconds
At that point,
v = at = 3.45*12 = 41.4 m/s
How long does it take for the canister to fall from that height, given its upward speed?
h(t) = 250 + 41.4t - 4.9t^2
h=0 after another 12.52 seconds
Now the rocket's height at that time is
h = 250 + 41.4t + 1.725 t^2
where t=12.52
h = 1038.72
To find the canister's maximum height, just find the vertex of
250 + 41.4t - 4.9t^2
That is 337.45m
so, it traveled 337.45*2 - 250 = 424.90 m after release.
As always, check my math.
1) The rocket and the fuel canister both experience the same acceleration due to gravity, which is approximately 9.8 m/s^2. Since the canister was disconnected when the rocket was 250 m above the launch pad, we can use the equation of motion:
h = ut + (1/2)at^2
where h is the final height, u is the initial velocity (0 m/s since it started from rest), a is the acceleration (-9.8 m/s^2), and t is the time taken.
By rearranging the equation, we get:
t = √(2h/a)
Substituting the values, we have:
t = √(2 * 250 / 9.8) ≈ √51.02 ≈ 7.14 s
Now, we can calculate the final height using the equation:
h = ut + (1/2)at^2
h = 0 * 7.14 + (1/2) * (-9.8) * (7.14)^2
h ≈ -248.6 m
Since the height cannot be negative, the rocket would be approximately 248.6 m above the launch pad when the canister hits the launch pad.
2) To find the total distance traveled by the canister, we need to calculate the distance it covers before hitting the launch pad. We can use the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity (0 m/s), a is the acceleration (-9.8 m/s^2), and t is the time taken.
Plugging in the values, we get:
s = 0 * 7.14 + (1/2) * (-9.8) * (7.14)^2
s ≈ -251.1 m
Again, since distance cannot be negative, the canister would have traveled approximately 251.1 m before crashing onto the launch pad.
To solve this problem, we can use the kinematic equations of motion. We will assume that the rocket starts from rest, so its initial velocity is 0 m/s.
1) To find the height of the rocket when the canister hits the launch pad, we can use the displacement formula:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
In this case, u = 0 m/s, a = 3.45 m/s^2, and we are given the displacement s = 250 m. We want to find the time it takes for the canister to hit the launch pad, so we rearrange the formula to solve for t:
t^2 = (2s) / a
t^2 = (2 * 250) / 3.45
t^2 = 145.65
t ≈ √145.65
t ≈ 12.08 s
Now we can use the time t to find the height of the rocket when the canister hits the launch pad:
s = ut + (1/2)at^2
s = 0 * 12.08 + (1/2) * 3.45 * (12.08)^2
s ≈ 248.99 m
Therefore, the height of the rocket when the canister hits the launch pad is approximately 248.99 m.
2) The total distance traveled by the canister is the sum of the distances traveled while accelerating upwards and while falling downwards. Both of these distances are equal.
The distance traveled while accelerating can be found using the formula:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
In this case, u = 0 m/s, a = 3.45 m/s^2, and t = 12.08 s (as found earlier).
s = 0 * 12.08 + (1/2) * 3.45 * (12.08)^2
s ≈ 248.99 m
Therefore, the distance traveled by the canister while accelerating upwards is approximately 248.99 m.
The distance traveled while falling downwards is also approximately 248.99 m.
Therefore, the total distance traveled by the canister between its release and its crash onto the launch pad is approximately 2 * 248.99 = 497.98 m.
To answer these questions, we can use the equations of motion. Let's start with question 1.
1) How high is the rocket when the canister hits the launch pad?
First, we need to find the time it takes for the canister to fall to the launch pad.
We know that the acceleration of the canister is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.
Using the equation of motion
s = ut + (1/2)at^2,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time, we can rearrange the equation to solve for time:
t = √(2s/a)
Plugging in the values:
s = 250 m (the height above the launch pad)
a = 9.8 m/s^2 (acceleration due to gravity)
t = √(2 * 250 / 9.8)
t ≈ √(51.0204)
t ≈ 7.1417 seconds
Now, we know the time it takes for the canister to fall. The rocket continues to accelerate at 3.45 m/s^2 during this time.
Using the equation of motion
s = ut + (1/2)at^2,
where s is the displacement of the rocket, u is the initial velocity (0 m/s as it starts from rest), a is the acceleration of the rocket (3.45 m/s^2), and t is the time (7.1417 seconds), we can calculate the displacement:
s = 0 + (1/2) * 3.45 * (7.1417)^2
s ≈ 164.61 m
Therefore, the rocket is approximately 164.61 meters above the launch pad when the canister hits the launch pad.
2) What total distance did the canister travel between its release and its crash onto the launch pad?
To find the total distance traveled by the canister, we need to calculate the distance it traveled while the rocket was accelerating and the distance it traveled while falling.
The distance traveled while the rocket was accelerating can be calculated using the equation of motion:
s = ut + (1/2)at^2
where:
u = 0 m/s (as the canister starts from rest)
a = 3.45 m/s^2
t = 7.1417 seconds (the time we calculated in the previous question)
s = 0 + (1/2) * 3.45 * (7.1417^2)
s ≈ 0 + (1/2) * 3.45 * 51.0204
s ≈ 88.057 m
The distance traveled while falling is equal to the height above the launch pad, which is 250 meters.
Therefore, the total distance traveled by the canister between its release and its crash onto the launch pad is approximately 88.057 + 250 = 338.057 meters.