An overly eager physics student wants to confirm that the acceleration of gravity is ag=9.8 m/s^2. To check this, the student steps off the roof of the Sears Tower (height at the sky deck 427m). The rocketeer (who arrived five seconds after the student stepped off) uses a rocket pack to accelerate downward in order to save the student.When the Rocketeer reaches the top of the tower he shuts off the rocket and then it is in free fall. Once the rocketeteer catches the student the rocket pack is used to accelerate both the students and rocketeer upwards until they come to a rest on the ground.The student can withstand a maximum acceleration of 5ag. What must the rocketeer's initial velocity at the top of the tower be in order to catch the student and at what height will the rocketeer catch the student?
At time t=x, the Rocketeer catches up with the student. At that point (the Rocketeer having arrived with some unknown velocity v), their height and speed are
student:
427 - 4.9x^2
9.8x
Rocketeer:
427 - v(x-5) - 4.9(x-5)^2
v + 9.8(x-5)
So, we match their heights and we get
427 - 4.9x^2 = 427 - v(x-5) - 4.9(x-5)^2
v = 49(x-2.5)/(x-5)
I see one problem here, in that there will be a might jerk when the Rocketeer grabs the student. We have to assume that at time t=x, the student's speed no longer matters, and both bodies are moving at the Rocketeer's speed.
At that time, then, taking y seconds at 5g rocket thrust to touch down, we have
427 - vy + 4*4.9y^2 = 0
v + 5*9.8y = 0
Work through that, and I get
v = 49(x-2.5)/(x-5) = 186.74
x = 5.88
y = 3.81
That is, after falling for 3.81 seconds, the Rocketeer catches the student at a height of
427 - 4.9*5.88^2 = 257.58m
The Rocketeer arrived at the top of the tower with a velocity of 186.74 m/s
Unfortunately, when the Rocketeer catches up with the student, he is going
186.74 + 9.8(5.88-5)-9.8*5.88 = 137.74 m/s
faster than the student. Poor student: the Rocketeer ran into him at a speed of 495 km/hr!
Maybe you should double-check my math... especially to make sure that they arrived at height=0 with speed=0
oops. After falling for 5.88 seconds, the student is caught, then 3.81 seconds later they land gently on the ground.
To find the initial velocity of the rocketeer at the top of the tower and the height at which they catch the student, we can use the equations of motion under constant acceleration.
Let's start by finding the time it takes for the student to reach the ground. We can use the equation of motion:
H = ut + (1/2)at^2
Where H is the height, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Height of the Sears Tower, H = 427 m
Acceleration due to gravity, ag = 9.8 m/s^2
We can solve this equation for time, t. Rearranging the equation, we get:
t = sqrt((2H) / a)
Substituting the values, we have:
t = sqrt((2 * 427) / 9.8)
t ≈ 9.92 seconds (rounded to two decimal places)
Now, let's determine how far the rocketeer falls during those 9.92 seconds. We can use the formula:
S = ut + (1/2)at^2
Since the rocketeer starts from rest at the top of the tower, the initial velocity, u, is 0. The acceleration, a, is the acceleration due to gravity, ag.
Therefore, the formula becomes:
S = (1/2)agt^2
Substituting the values, we have:
S = (1/2) * 9.8 * (9.92)^2
S ≈ 485.66 meters (rounded to two decimal places)
This distance represents the height the rocketeer falls during those 9.92 seconds.
Since the rocketeer needs to catch the student, who falls for the same amount of time, they must start accelerating upwards from this height.
Now we need to use the equation of motion to find the initial velocity, u, of the rocketeer at the top of the tower, for the rocketeer to reach the student in time.
Let's denote the distance at which the rocketeer catches the student as x. The initial velocity, u, of the rocketeer can be found using the equation:
x = ut + (1/2)at^2
Since the student falls for 9.92 seconds and the rocketeer starts from rest, the equation becomes:
x = (1/2)au^2
But we also know that the student can withstand a maximum acceleration of 5ag.
Therefore, the maximum acceleration, amax, the student experiences is:
amax = 5ag = 5 * 9.8 = 49 m/s^2
Now, equating the maximum acceleration of the student to the acceleration experienced by the rocketeer, we have:
amax = (1/2)au^2
Substituting the value for amax, we get:
49 = (1/2) * 9.8 * u^2
Simplifying, we have:
u^2 = (49 * 2) / 9.8
u^2 ≈ 10
Taking the square root, we get:
u ≈ sqrt(10)
u ≈ 3.16 m/s (rounded to two decimal places)
Finally, to find the height at which the rocketeer catches the student, we need to find the distance fallen by the rocketeer during those 9.92 seconds.
Using the equation of motion:
S = ut + (1/2)at^2
Where S is the distance fallen by the rocketeer, u is the initial velocity, a is the acceleration, and t is the time.
Since the initial velocity, u, is 3.16 m/s (as calculated above), the time, t, is 9.92 seconds, and the acceleration, a, is 9.8 m/s^2, we have:
S = (3.16 * 9.92) + (1/2) * 9.8 * (9.92)^2
S ≈ 198.55 meters (rounded to two decimal places)
Therefore, the height at which the rocketeer catches the student is approximately 198.55 meters.
To summarize:
- The initial velocity of the rocketeer at the top of the tower should be approximately 3.16 m/s.
- The rocketeer catches the student at a height of approximately 198.55 meters.