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?
2 answers
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