Finding the average height and average velocity of a ball dropped from height of $400$ ft












2












$begingroup$



Rosanne drops a ball from a height of 400 ft. Find the ball's
average height and its average velocity between the time it is
dropped and the time it strikes the ground.




My trial...



So, I tried to use average value theorem for integrals. I took acceleration as positive $32$ ft m/s$^2$ (as velocity increases). By integrating it, I found velocity as $v=32t + C$ (with $C=0$). I integrated it one more time and got the distance $16 t^2 = 400$. And now $t = 5$ sec. After that, I tried to use average value of the function theorem. Somehow got the wrong answer. For finding H average and I did the following. (1/5)*(integral(from 0 to 5) Htdt. For this integral I found the answer 400/3 while answer given is 800/3






calculus integration







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  • 2




    $begingroup$
    I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
    $endgroup$
    – Saad
    Dec 17 '18 at 0:52






  • 2




    $begingroup$
    @user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
    $endgroup$
    – gimusi
    Dec 18 '18 at 7:21
















2












$begingroup$



Rosanne drops a ball from a height of 400 ft. Find the ball's
average height and its average velocity between the time it is
dropped and the time it strikes the ground.




My trial...



So, I tried to use average value theorem for integrals. I took acceleration as positive $32$ ft m/s$^2$ (as velocity increases). By integrating it, I found velocity as $v=32t + C$ (with $C=0$). I integrated it one more time and got the distance $16 t^2 = 400$. And now $t = 5$ sec. After that, I tried to use average value of the function theorem. Somehow got the wrong answer. For finding H average and I did the following. (1/5)*(integral(from 0 to 5) Htdt. For this integral I found the answer 400/3 while answer given is 800/3






calculus integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
    $endgroup$
    – Saad
    Dec 17 '18 at 0:52






  • 2




    $begingroup$
    @user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
    $endgroup$
    – gimusi
    Dec 18 '18 at 7:21














2












2








2





$begingroup$



Rosanne drops a ball from a height of 400 ft. Find the ball's
average height and its average velocity between the time it is
dropped and the time it strikes the ground.




My trial...



So, I tried to use average value theorem for integrals. I took acceleration as positive $32$ ft m/s$^2$ (as velocity increases). By integrating it, I found velocity as $v=32t + C$ (with $C=0$). I integrated it one more time and got the distance $16 t^2 = 400$. And now $t = 5$ sec. After that, I tried to use average value of the function theorem. Somehow got the wrong answer. For finding H average and I did the following. (1/5)*(integral(from 0 to 5) Htdt. For this integral I found the answer 400/3 while answer given is 800/3






calculus integration







share|cite|improve this question











$endgroup$





Rosanne drops a ball from a height of 400 ft. Find the ball's
average height and its average velocity between the time it is
dropped and the time it strikes the ground.




My trial...



So, I tried to use average value theorem for integrals. I took acceleration as positive $32$ ft m/s$^2$ (as velocity increases). By integrating it, I found velocity as $v=32t + C$ (with $C=0$). I integrated it one more time and got the distance $16 t^2 = 400$. And now $t = 5$ sec. After that, I tried to use average value of the function theorem. Somehow got the wrong answer. For finding H average and I did the following. (1/5)*(integral(from 0 to 5) Htdt. For this integral I found the answer 400/3 while answer given is 800/3






calculus integration




calculus integration






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 10:43







Arif Rustamov

















asked Dec 16 '18 at 10:10









Arif RustamovArif Rustamov

367




367








  • 2




    $begingroup$
    I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
    $endgroup$
    – Saad
    Dec 17 '18 at 0:52






  • 2




    $begingroup$
    @user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
    $endgroup$
    – gimusi
    Dec 18 '18 at 7:21














  • 2




    $begingroup$
    I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
    $endgroup$
    – Saad
    Dec 17 '18 at 0:52






  • 2




    $begingroup$
    @user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
    $endgroup$
    – gimusi
    Dec 18 '18 at 7:21








2




2




$begingroup$
I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
$endgroup$
– Saad
Dec 17 '18 at 0:52




$begingroup$
I'm voting to close this question as off-topic because the "trial" was added after the almost complete solution has been posted for 10+ minutes.
$endgroup$
– Saad
Dec 17 '18 at 0:52




2




2




$begingroup$
@user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
$endgroup$
– gimusi
Dec 18 '18 at 7:21




$begingroup$
@user302797 Note that at first the aker posted his work by comments now deleted and it was done very quickly. Please reviseyour evaluation. Thanks
$endgroup$
– gimusi
Dec 18 '18 at 7:21










2 Answers
2






active

oldest

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0












$begingroup$

Assuming $v_0=0$, we have that




  • $h(t)=400-frac12gt^2 implies t_{max}=sqrt{frac{800}{g}}=20sqrt{frac{2}{g}}$ time to strike the ground


  • $v(t)=gt implies v_{max}=20sqrt{2g} implies bar v=frac{v_0+v_{max}}{2}$



and



$$bar h = frac{int_0^{t_{max}}h(t) dt}{t_{max}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:30












  • $begingroup$
    @SameerBaheti It holds because $v$ is linear in that case.
    $endgroup$
    – gimusi
    Dec 16 '18 at 12:32










  • $begingroup$
    Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:38



















0












$begingroup$

Let $H=400text{ft}$.
$$h(t)=H-frac12gt^2$$
Average height = particular height $times$ time during which it was on that height/ total time.
$$frac12gT^2=Himplies T=sqrt{frac{2H}g}$$
Or mathematically
$$<h>=frac{int_0^Th(t)dt}{int_0^Tdt}$$
$$<h>=frac{int_0^T(H-frac12gt^2)dt}T=H-frac{frac16gbig[t^3big]_0^T}T=H-frac{H}3=frac{2H}3=frac{800}3text{ft}$$
$v(t)=gt.$
So, similarly $$<v>=frac1Tint_0^Tgtdt=frac HT=sqrt{frac{Hg}2}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    0












    $begingroup$

    Assuming $v_0=0$, we have that




    • $h(t)=400-frac12gt^2 implies t_{max}=sqrt{frac{800}{g}}=20sqrt{frac{2}{g}}$ time to strike the ground


    • $v(t)=gt implies v_{max}=20sqrt{2g} implies bar v=frac{v_0+v_{max}}{2}$



    and



    $$bar h = frac{int_0^{t_{max}}h(t) dt}{t_{max}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:30












    • $begingroup$
      @SameerBaheti It holds because $v$ is linear in that case.
      $endgroup$
      – gimusi
      Dec 16 '18 at 12:32










    • $begingroup$
      Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:38
















    0












    $begingroup$

    Assuming $v_0=0$, we have that




    • $h(t)=400-frac12gt^2 implies t_{max}=sqrt{frac{800}{g}}=20sqrt{frac{2}{g}}$ time to strike the ground


    • $v(t)=gt implies v_{max}=20sqrt{2g} implies bar v=frac{v_0+v_{max}}{2}$



    and



    $$bar h = frac{int_0^{t_{max}}h(t) dt}{t_{max}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:30












    • $begingroup$
      @SameerBaheti It holds because $v$ is linear in that case.
      $endgroup$
      – gimusi
      Dec 16 '18 at 12:32










    • $begingroup$
      Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:38














    0












    0








    0





    $begingroup$

    Assuming $v_0=0$, we have that




    • $h(t)=400-frac12gt^2 implies t_{max}=sqrt{frac{800}{g}}=20sqrt{frac{2}{g}}$ time to strike the ground


    • $v(t)=gt implies v_{max}=20sqrt{2g} implies bar v=frac{v_0+v_{max}}{2}$



    and



    $$bar h = frac{int_0^{t_{max}}h(t) dt}{t_{max}}$$






    share|cite|improve this answer









    $endgroup$



    Assuming $v_0=0$, we have that




    • $h(t)=400-frac12gt^2 implies t_{max}=sqrt{frac{800}{g}}=20sqrt{frac{2}{g}}$ time to strike the ground


    • $v(t)=gt implies v_{max}=20sqrt{2g} implies bar v=frac{v_0+v_{max}}{2}$



    and



    $$bar h = frac{int_0^{t_{max}}h(t) dt}{t_{max}}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 16 '18 at 10:27









    gimusigimusi

    92.8k84494




    92.8k84494












    • $begingroup$
      Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:30












    • $begingroup$
      @SameerBaheti It holds because $v$ is linear in that case.
      $endgroup$
      – gimusi
      Dec 16 '18 at 12:32










    • $begingroup$
      Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:38


















    • $begingroup$
      Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:30












    • $begingroup$
      @SameerBaheti It holds because $v$ is linear in that case.
      $endgroup$
      – gimusi
      Dec 16 '18 at 12:32










    • $begingroup$
      Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
      $endgroup$
      – Sameer Baheti
      Dec 16 '18 at 12:38
















    $begingroup$
    Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:30






    $begingroup$
    Does this always hold $bar v=frac{v_0+v_{max}}{2}$? I mean it is good in this case but it should be total displacement/ total time.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:30














    $begingroup$
    @SameerBaheti It holds because $v$ is linear in that case.
    $endgroup$
    – gimusi
    Dec 16 '18 at 12:32




    $begingroup$
    @SameerBaheti It holds because $v$ is linear in that case.
    $endgroup$
    – gimusi
    Dec 16 '18 at 12:32












    $begingroup$
    Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:38




    $begingroup$
    Oh yes! If acceleration is constant and path is straight line one can prove this by $v^2-u^2=2as$.
    $endgroup$
    – Sameer Baheti
    Dec 16 '18 at 12:38











    0












    $begingroup$

    Let $H=400text{ft}$.
    $$h(t)=H-frac12gt^2$$
    Average height = particular height $times$ time during which it was on that height/ total time.
    $$frac12gT^2=Himplies T=sqrt{frac{2H}g}$$
    Or mathematically
    $$<h>=frac{int_0^Th(t)dt}{int_0^Tdt}$$
    $$<h>=frac{int_0^T(H-frac12gt^2)dt}T=H-frac{frac16gbig[t^3big]_0^T}T=H-frac{H}3=frac{2H}3=frac{800}3text{ft}$$
    $v(t)=gt.$
    So, similarly $$<v>=frac1Tint_0^Tgtdt=frac HT=sqrt{frac{Hg}2}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $H=400text{ft}$.
      $$h(t)=H-frac12gt^2$$
      Average height = particular height $times$ time during which it was on that height/ total time.
      $$frac12gT^2=Himplies T=sqrt{frac{2H}g}$$
      Or mathematically
      $$<h>=frac{int_0^Th(t)dt}{int_0^Tdt}$$
      $$<h>=frac{int_0^T(H-frac12gt^2)dt}T=H-frac{frac16gbig[t^3big]_0^T}T=H-frac{H}3=frac{2H}3=frac{800}3text{ft}$$
      $v(t)=gt.$
      So, similarly $$<v>=frac1Tint_0^Tgtdt=frac HT=sqrt{frac{Hg}2}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $H=400text{ft}$.
        $$h(t)=H-frac12gt^2$$
        Average height = particular height $times$ time during which it was on that height/ total time.
        $$frac12gT^2=Himplies T=sqrt{frac{2H}g}$$
        Or mathematically
        $$<h>=frac{int_0^Th(t)dt}{int_0^Tdt}$$
        $$<h>=frac{int_0^T(H-frac12gt^2)dt}T=H-frac{frac16gbig[t^3big]_0^T}T=H-frac{H}3=frac{2H}3=frac{800}3text{ft}$$
        $v(t)=gt.$
        So, similarly $$<v>=frac1Tint_0^Tgtdt=frac HT=sqrt{frac{Hg}2}$$






        share|cite|improve this answer









        $endgroup$



        Let $H=400text{ft}$.
        $$h(t)=H-frac12gt^2$$
        Average height = particular height $times$ time during which it was on that height/ total time.
        $$frac12gT^2=Himplies T=sqrt{frac{2H}g}$$
        Or mathematically
        $$<h>=frac{int_0^Th(t)dt}{int_0^Tdt}$$
        $$<h>=frac{int_0^T(H-frac12gt^2)dt}T=H-frac{frac16gbig[t^3big]_0^T}T=H-frac{H}3=frac{2H}3=frac{800}3text{ft}$$
        $v(t)=gt.$
        So, similarly $$<v>=frac1Tint_0^Tgtdt=frac HT=sqrt{frac{Hg}2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 12:22









        Sameer BahetiSameer Baheti

        5168




        5168






























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