mastering physics
posted by merve .
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\left( t \right) = b  ct + dt^2, where b = 740 m is the initial height of the lander above the surface, c = 64.0 m/s, and d = 1.03 m/s^2.What is the velocity of the lander just before it reaches the lunar surface? please answer quikly.

mastering physics 
Muzna
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 nonphysical 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..!)
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