A question about free fall, velocity, and the height of an object.












4












$begingroup$


A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.



a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?



b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?



I got the wrong answers when working on this.



To solve a):



$$s(t+2) - s(t) = 84$$
$$s(t) = v_0t+cfrac{1}{2}at^2, v_0 = 5, a = 32$$
$$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
$$64t=10$$
$$t=cfrac{5}{8}$$
$$5left(cfrac{5}{8}right)+16left(cfrac{5}{8}right)^2=9.375$$
$$h_0=109.375$$



To solve b):



$$100=-16t^2+7t+h_0$$
$$16=-16(t+2)^2+7(t+2)+h_0$$
now subtract the smaller constant from the larger
$$-84=-71t+7t-50$$
$$t=cfrac{34}{71}$$
$$100=-16left(cfrac{34}{71}right)^2+7left(cfrac{34}{71}right)+h_0$$
$$h_0=cfrac{505698}{5041}$$



However the answers are:
$a=cfrac{6475}{65}$
$b=100$



What am I doing wrong?






calculus







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.



    a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?



    b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?



    I got the wrong answers when working on this.



    To solve a):



    $$s(t+2) - s(t) = 84$$
    $$s(t) = v_0t+cfrac{1}{2}at^2, v_0 = 5, a = 32$$
    $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
    $$64t=10$$
    $$t=cfrac{5}{8}$$
    $$5left(cfrac{5}{8}right)+16left(cfrac{5}{8}right)^2=9.375$$
    $$h_0=109.375$$



    To solve b):



    $$100=-16t^2+7t+h_0$$
    $$16=-16(t+2)^2+7(t+2)+h_0$$
    now subtract the smaller constant from the larger
    $$-84=-71t+7t-50$$
    $$t=cfrac{34}{71}$$
    $$100=-16left(cfrac{34}{71}right)^2+7left(cfrac{34}{71}right)+h_0$$
    $$h_0=cfrac{505698}{5041}$$



    However the answers are:
    $a=cfrac{6475}{65}$
    $b=100$



    What am I doing wrong?






    calculus







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.



      a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?



      b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?



      I got the wrong answers when working on this.



      To solve a):



      $$s(t+2) - s(t) = 84$$
      $$s(t) = v_0t+cfrac{1}{2}at^2, v_0 = 5, a = 32$$
      $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
      $$64t=10$$
      $$t=cfrac{5}{8}$$
      $$5left(cfrac{5}{8}right)+16left(cfrac{5}{8}right)^2=9.375$$
      $$h_0=109.375$$



      To solve b):



      $$100=-16t^2+7t+h_0$$
      $$16=-16(t+2)^2+7(t+2)+h_0$$
      now subtract the smaller constant from the larger
      $$-84=-71t+7t-50$$
      $$t=cfrac{34}{71}$$
      $$100=-16left(cfrac{34}{71}right)^2+7left(cfrac{34}{71}right)+h_0$$
      $$h_0=cfrac{505698}{5041}$$



      However the answers are:
      $a=cfrac{6475}{65}$
      $b=100$



      What am I doing wrong?






      calculus







      share|cite|improve this question









      $endgroup$




      A falling stone is at a certain instant $100$ feet above the ground. Two seconds later it is only $16$ feet above the ground.



      a) If it was thrown downward with an initial speed of $5$ ft/sec, from what height was it thrown?



      b) If it was thrown upward with an initial speed of $10$ ft/sec, from what height was it thrown?



      I got the wrong answers when working on this.



      To solve a):



      $$s(t+2) - s(t) = 84$$
      $$s(t) = v_0t+cfrac{1}{2}at^2, v_0 = 5, a = 32$$
      $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
      $$64t=10$$
      $$t=cfrac{5}{8}$$
      $$5left(cfrac{5}{8}right)+16left(cfrac{5}{8}right)^2=9.375$$
      $$h_0=109.375$$



      To solve b):



      $$100=-16t^2+7t+h_0$$
      $$16=-16(t+2)^2+7(t+2)+h_0$$
      now subtract the smaller constant from the larger
      $$-84=-71t+7t-50$$
      $$t=cfrac{34}{71}$$
      $$100=-16left(cfrac{34}{71}right)^2+7left(cfrac{34}{71}right)+h_0$$
      $$h_0=cfrac{505698}{5041}$$



      However the answers are:
      $a=cfrac{6475}{65}$
      $b=100$



      What am I doing wrong?






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      JinzuJinzu

      403513




      403513






















          2 Answers
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          1












          $begingroup$

          The error in a) is simple:



          From $64t=10$ it follows $t=frac5{32} neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:



          $h_0=100+5left(frac58right) + 16left(frac58right)^2=frac{6475}{64}$



          which is very similar to your answer key (I assume you mistyped the denominator).



          In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            the solution of
            $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
            should be $t=frac{5}{32}$ not $t=frac{5}{8}$






            share|cite|improve this answer









            $endgroup$














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              2 Answers
              2






              active

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              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The error in a) is simple:



              From $64t=10$ it follows $t=frac5{32} neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:



              $h_0=100+5left(frac58right) + 16left(frac58right)^2=frac{6475}{64}$



              which is very similar to your answer key (I assume you mistyped the denominator).



              In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The error in a) is simple:



                From $64t=10$ it follows $t=frac5{32} neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:



                $h_0=100+5left(frac58right) + 16left(frac58right)^2=frac{6475}{64}$



                which is very similar to your answer key (I assume you mistyped the denominator).



                In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The error in a) is simple:



                  From $64t=10$ it follows $t=frac5{32} neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:



                  $h_0=100+5left(frac58right) + 16left(frac58right)^2=frac{6475}{64}$



                  which is very similar to your answer key (I assume you mistyped the denominator).



                  In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.






                  share|cite|improve this answer









                  $endgroup$



                  The error in a) is simple:



                  From $64t=10$ it follows $t=frac5{32} neq frac58$. Substituting this into your formula for $s(t)$ (including that after time $t$ you are at $100$ft) yields:



                  $h_0=100+5left(frac58right) + 16left(frac58right)^2=frac{6475}{64}$



                  which is very similar to your answer key (I assume you mistyped the denominator).



                  In b) you seem to be calculating with $v_0=7ft/s$, but $v_0=10ft/s$ was given.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  IngixIngix

                  5,097159




                  5,097159























                      1












                      $begingroup$

                      the solution of
                      $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
                      should be $t=frac{5}{32}$ not $t=frac{5}{8}$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        the solution of
                        $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
                        should be $t=frac{5}{32}$ not $t=frac{5}{8}$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          the solution of
                          $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
                          should be $t=frac{5}{32}$ not $t=frac{5}{8}$






                          share|cite|improve this answer









                          $endgroup$



                          the solution of
                          $$left[5(t+2)+16(t+2)^2right]-(5t+16t^2)=84$$
                          should be $t=frac{5}{32}$ not $t=frac{5}{8}$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 4 hours ago









                          E.H.EE.H.E

                          16.1k11968




                          16.1k11968






























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