Object falling due to Gravity Problem

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Homework Help Overview

The problem involves a lander descending to the moon's surface, where it experiences free fall after its engine is cut off. The initial conditions include a height of 5.0 m above the surface and a downward speed of 0.8 m/s, with the moon's gravitational acceleration given as 1.6 m/s². The goal is to determine the speed of the lander just before it touches the surface.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the appropriate kinematic equations to use for calculating the final speed of the lander. There are questions about the correct application of signs in the equations and the implications of negative results when calculating speed.

Discussion Status

The discussion has seen participants attempting to apply kinematic equations, with some guidance provided regarding the correct formulation and use of signs. There is an ongoing exploration of the implications of their calculations, particularly concerning the signs and the resulting values.

Contextual Notes

Participants are navigating issues related to the choice of positive and negative signs in their calculations, which affects their results. There is also a mention of potential confusion arising from the square root of negative values in their equations.

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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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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.
 
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.
 
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.
 
Okay but now what do I do becase I would have a square root of a negative which is not possible?
 
Last edited:
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.
 
mopar969 said:
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) = +[/color] 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:
 
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.
 
mopar969 said:
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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