Show how to do part e ht D: Fly the lander so that it is flying straight up with

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Show how to do part e ht D: Fly the lander so that it is flying straight up with vy>25 m/s. Tun off the thruster so the have free fall. (There is no atmosphere on the moon so this is a realistic free fall situation). Shortly after it is in free fall, values w start the timer as you crashes. Record the values of the times, the max height, and vy just before crashing. (I think the crash speed is displayed after the crash and since it was going straight down, vy should be the negative of the start a stopwatch and at the same time measure vy and the altitude y. These are the t-0. (Unless you have a co-pilot, you may want to pause the flight to take the readings and long it takes before the lander reaches speed.) Q2.1a) when t-o voy"-moll - At maximum heighty At the crash:y b) Use the definition of average acceleration to calculate the value of the acceleration for the (from t-0 until the crash). acceleration is constant. Since there is only a constant weight force we are assuming that the 70 e) Think about the free body diagram for an arbitrary object during free fall. What forces are there? Using the fact that the magnitude of the weight is mg and its direction is down, write an expression for the y- component ofthe weight (in the coordinate system with +y up). Then idea that Fy-ma, to write an expression for ay for an arbitrary object in casily use your value of a, for your test flight to solve for "g" of the moon free fall (weight force only). At this point you should be able to Formule foray for an arbitrary object in free fall: ay=- Value of"g" for the moon based on test flight D d) Write the motion equations for a trip with constant acceleration along the y-direction for the test flight from when you started the timer until the crash. These should be as functions of time alone. For the whole trip 2aali-h44 mete y- alentamist-IH6NA'Y vy- e) On separate paper use the motion equations to predict the time or the maximum height and the max height and the time and speed of the crash. (Show work on separate paper and attach it but put answers here) Predicted max height: time- height = " . Predicted crash: time = speed .

Explanation / Answer

Part e. as we can see
   at t = 0, vy = 29 m/s, yo = 291 m ( from part 1)
   so at max height
   let the time be t
   then at max height vy = 0
   so 0 = vy - gt
   t = vy/g
   g for moon = 1.64 m/s/s
   hence t = 29/1.64 = 17.682 s
   and maximum heigt = H + yo
   then
   2*g*H = vy^2
   2*1.64*H = 29^2
   H = 256.4024 m
   hence max height = H + yo = 547.402439

   now, predicted crash time t= t'
   then
   -yo = vyt - 0.5gt^2
   -291 = 29*t' - 0.5*1.64*t'^2
   0.82*t'^2 - 29t' - 291 = 0
   solving for t'
   t' = 43.52 s

   crash speed = v
   then
   v = vy - gt'
   v = 29 - 1.64*43.52
   v = -42.373 m/s

   so crash speed = 42.373 m/s

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