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







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












$begingroup$



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






share|cite|improve this question















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

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042433%2ffinding-the-average-height-and-average-velocity-of-a-ball-dropped-from-height-of%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042433%2ffinding-the-average-height-and-average-velocity-of-a-ball-dropped-from-height-of%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more