Solving for Lunar Landing Craft Thrust

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Homework Statement



A 8.73 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is 1.60 m/s2. At an altitude of 217 m the craft's downward velocity is 16.5 m/s. To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c04/EAT_12257263467150_5449420724903514.gif

Homework Equations



v^2=v0^2+2ax

The Attempt at a Solution



Fthrust=Fg? mg=ma?
 
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Hello onyxorca,

As a first step you have to figure out which forces act on the rocket. Then you have to use the general equations for x(t) and v(t).
 
F=ma

the forces are gravity and thrust.

btw the mass is 8.73 × 10^3 kg.

the acceleration is 0=16.5^2+2a217 and a = -0.627304 m/s^2 ? where do i go from there?
 
Fthrust - Fg = ma?

so F - 8730*1.6 = 8730*-0.627304

F=8491.64 N?
 
F_thrust -F_g = ma is correct! :smile:

Now you need to find out the value of a.

For that use the general equations for x(t) and v(t):

v(t) = a*t + v_0

x(t) = 1/2*a*t^2 + v_0*t + x_0
 
well appearantly 8491 isn't right...
 
V0=V^2+2ax find a

F=mg-ma=m(g-a)

i finally got it right, had trouble realizing that the thrust is smaller than gravitational force and i made mistakes when putting in the scientific digits but i finally got it.
 
well the numbers in the problem changes every time so i wouldn't know if this is right but i got the right answer for that time.

in this case though it's probably F = 8.73E3 (1.6+.627304) = 19444.4 N because it's Fg-Fthrust, not the other way around.
 
I get the same number.