University of Houston

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Problem 63

Lunar Gravity On the moon, the acceleration of a free-falling object is $a(t)=-1.6$ meters per second per

second. A stone is dropped from a cliff on the moon and hits the surface of the moon 20 seconds later. How far did it fall? What was its velocity at impact?

Answer

$$

-32 \mathrm{m} / \mathrm{s}

$$

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## Discussion

## Video Transcript

Eric. Question three we have he negative. How far did it fall? That tiny 20 and what was lost? The on impact Remember a t eat booms as double prime. So to find s prime you take the inner integral of negative 1.6 dp which gives me negative 1.6 he plus C which in this case is our initial velocity, which is zero. So I don't really have a plus. So my guess Crime negative 116 and my position function When I take the integral again one t do teen is gonna give you Oh, negative 1.6 She's squared and multiply that and the initial height And she's what we're trying to find from it Initial c So my s a T equals negative 0.5 times and they have six Gives me negative point piece We're close my initial height So now at 20 at time equals 20. Well, give me my initial height. There weird condition make It is my mission. So see equal 320 And that would be in meters. And then when I want to find out the loss of the onion, I'm gonna go to my bride s t negative. 1.6. I'm sweating. Actions would be a friend. And that's a negative. There it is, my velocity.

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