A probe is ejected upward from a lunar module 3000 ft above the surface of the moon.?

After 10 seconds, the probe is 2980 ft above the lunar surface. After 20 seconds, it is 2420 feet above the lunar surface.





A) Determine the position function for the distance of the probe above the surface at any time





B) How many seconds will it take for the probe to reach the lunar surface?|||A)


at t=0, h = 3000


at t=10, h = 2980


at t=20, h = 2420





in the 1st 10 sec, the probe fell 20 ft


in the 2nd 10 sec, the probe fell 560 ft





Let us assume that the probe is in a parabolic fall. Therefore, the function for the height is





f(t) = a1*t^2 + a2*t + a3





at


f(0) = 0 + 0 + a3 = 3000





So a3 = 3000


Then,


f(10) = a1*(100) + a2*10 + 3000 = 2980


f(20) = a1*(400) + a2*20 + 3000 = 2420





Using f(10) and f(20) we have 2 equations with 2 unkowns, a1 and a2. Reducing them further yeilds:





f(10) = 100*a1 + 10*a2 = -20


f(20) = 400*a1 + 20*a2 = -580





I'll solve this using matricies. If you need another method, you can do that by hand...





A =


[100 10]


[400 20]





b =


[-20 ]


[-580]





x =


[a1]


[a2]





A*x = b


x = A^-1*b





x =


[-2.7]


[25.0]





so





f(t) = -2.7*t^2 + 25*t + 3000





B)


Set f(t) = 0 and solve for t


-2.7*t^2 + 25*t + 3000 = 0


t = 38.3 sec