Object falling due to Gravity Problem

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  • #1
mopar969
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A lander is descending to land on the moon. The engine is cut off when the lander is 5.0 m above the surface and has a downward speed of 0.8 m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 m/s^2. Please help me start with this problem thanks.
 

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  • #2
PhanthomJay
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When an object is in free fall, only the gravity force is acting on it. So what is its downward acceleration? Use one of the kinematic equations to solve for its speed.
 
  • #3
mopar969
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I used the equation V = V0 + 2a(y-y0)
= (0.8)+2(1.6)(0-5)
= -15.2 m/s

Please check and make sure that I used the right equation and obtained the correct answer.
 
  • #4
PhanthomJay
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Correct your equation...check units...it's v^2 = etc...
Also, be careful of plus and minus signs ...since you chose down as positive for initial velocity and acceleration, then the displacement down is also positive.
 
  • #5
mopar969
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Okay but now what do I do becase I would have a square root of a negative which is not possible?
 
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  • #6
PhanthomJay
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Well, look it up...it's v_f^2 = v_o^2 + 2a(delta y), for constant acceleration...or you can derive it from the calculus...best to memorize it, once you know its derivation.
 
  • #7
PhanthomJay
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Okay but now what do I do becase I would have a square root of a negative which is not possible?
Again, watch your plus and minus signs..
it's v^2 =v_o^2 + 2g(y - y_o); Since you chose down as positive, then (y - y_o) is (5- 0) = + 5. If you chose up as positive, then its v^2 = (-V_o)^2 + 2(-g)(-5 -0), which yields the same result. Those plus and minus signs will try to get you every time..don't let them.:wink:
 
  • #8
mopar969
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Thanks for the negative sign help. After I fixed everything I obtained an answer of 4.08 m/s. Can you tell me if my answer is correct. Thanks.
 
  • #9
Daever
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Thanks for the negative sign help. After I fixed everything I obtained an answer of 4.08 m/s. Can you tell me if my answer is correct. Thanks.

it is correct.
 

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