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1-D motion freefall

  1. Oct 21, 2013 #1
    1. At t=0, a stone is dropped from the top of a cliff above a lake. Another stone is thrown downward 1.6 s later from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff.



    2. d[itex]_{1}[/itex]=[itex]\frac{1}{2}[/itex]gt[itex]_{1}[/itex][itex]^{2}[/itex]; d[itex]_{2}[/itex]=v[itex]_{02}[/itex]t[itex]_{2}[/itex]+[itex]\frac{1}{2}[/itex]gt[itex]_{2}[/itex][itex]^{2}[/itex];d[itex]_{1}[/itex]=d[itex]_{2}[/itex]; t[itex]_{2}[/itex]=t[itex]_{1}[/itex]- 1.6s



    3. [itex]\frac{1}{2}[/itex]gt[itex]_{1}[/itex][itex]^{2}[/itex]=v[itex]_{02}[/itex]t[itex]_{2}[/itex]+[itex]\frac{1}{2}[/itex]g(t[itex]_{1}[/itex]-1.6s)[itex]^{2}[/itex]. [itex]\frac{1}{2}[/itex](9.81 m/s[itex]^{2}[/itex])t[itex]_{1}[/itex][itex]^{2}[/itex]=(32 m/s)(t[itex]_{1}[/itex]-1.6s)+[itex]\frac{1}{2}[/itex](9.81 m/s[itex]^{2}[/itex])(t[itex]_{1}[/itex]-1.6s)[itex]^{2}[/itex]. Here is where I get stuck in solving the quadratic.
     
  2. jcsd
  3. Oct 21, 2013 #2
    You should solve the equation yourself as it contains only maths. The equation will give you two values and i think one value of t1 will come negative which you must ignore.Try it :)
     
  4. Oct 21, 2013 #3
    For some reason I get 2.82 seconds for t.
     
  5. Oct 21, 2013 #4
    According to my calculation t comes to be 2.02 second
     
  6. Oct 21, 2013 #5
    So height of cliff is 20m
     
  7. Oct 21, 2013 #6
    Note i have taken g=10m/s^2
     
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