Posted by merve on Sunday, October 16, 2011 at 3:39am.
A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by y(t) = b - ct + dt^2, where b = 750 m is the initial height of the lander above the surface, c = 65.0 m/s, and d = 1.03 m/s^2.
What is the velocity of the lander just before it reaches the lunar surface?
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Best Answer
Dr Zorro answered 3 years ago
The time when the lunar lander reaches the surface is found by solving y(t)=0.
b - ct + d t^2 = 0 => t = (c +- sqrt( c^2 - 4 b d)) / (2d)
t = (65.0 m/s +- sqrt((65.0m/s)^2 - 4 * 750 m * 1.03 m/s^2))/(2*1.03 m/s^2)
t = (65.0 m/s +- 33.7 m/s) / (2.6 m/s^2)
t = 12.0 s or t = 38.0 s .
Obviously we need the smallest time solution (when y equals zero from the first time, the second solution being non-physical as it is the solution corresponding to the lunar lander going down through the surface and coming back up to y=0 from under the surface)
So y=0 when t=12.0 s
The velocity can now be found from:
v(t) = dy/dt = -c + 2 d t
So at the surface we have
v(12.0s) = -65.0 m/s + 2 * 1.03 m/s^2 * 12.0s
= -40.3 m/s
I.e. a downward speed of 40.3 m/s (Although correctly calculated, this seems a bit large to me as it should be a lunar lander, not a lunar crasher..!)