Find the initial direction and time of flight of a basketball, given initial speed and distance












0












$begingroup$



A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?




I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.






algebra-precalculus trigonometry physics







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$endgroup$












  • $begingroup$
    Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:13












  • $begingroup$
    I didn't understand that, what's the latter interperatation
    $endgroup$
    – Zayn Malek
    Oct 18 '15 at 0:17










  • $begingroup$
    It's what @C.I.J. assumes in his solution
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:20


















0












$begingroup$



A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?




I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.






algebra-precalculus trigonometry physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:13












  • $begingroup$
    I didn't understand that, what's the latter interperatation
    $endgroup$
    – Zayn Malek
    Oct 18 '15 at 0:17










  • $begingroup$
    It's what @C.I.J. assumes in his solution
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:20
















0












0








0





$begingroup$



A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?




I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.






algebra-precalculus trigonometry physics







share|cite|improve this question











$endgroup$





A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?




I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.






algebra-precalculus trigonometry physics




algebra-precalculus trigonometry physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 18 '15 at 5:06







user147263

















asked Oct 17 '15 at 23:47









Zayn MalekZayn Malek

63




63












  • $begingroup$
    Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:13












  • $begingroup$
    I didn't understand that, what's the latter interperatation
    $endgroup$
    – Zayn Malek
    Oct 18 '15 at 0:17










  • $begingroup$
    It's what @C.I.J. assumes in his solution
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:20




















  • $begingroup$
    Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:13












  • $begingroup$
    I didn't understand that, what's the latter interperatation
    $endgroup$
    – Zayn Malek
    Oct 18 '15 at 0:17










  • $begingroup$
    It's what @C.I.J. assumes in his solution
    $endgroup$
    – Semiclassical
    Oct 18 '15 at 0:20


















$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13






$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13














$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17




$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17












$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20






$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20












2 Answers
2






active

oldest

votes


















0












$begingroup$

If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is



    $$ frac{distance }{v cos alpha} $$



    So $alpha$ should also be known.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The theta wasn't given.
      $endgroup$
      – Zayn Malek
      Oct 18 '15 at 0:05










    • $begingroup$
      If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
      $endgroup$
      – Narasimham
      Oct 18 '15 at 0:09












    • $begingroup$
      Ooh I get that. How did you derive that equations
      $endgroup$
      – Zayn Malek
      Oct 18 '15 at 0:13










    • $begingroup$
      Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
      $endgroup$
      – Narasimham
      Oct 18 '15 at 0:32











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.






        share|cite|improve this answer









        $endgroup$



        If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 18 '15 at 0:20









        BatominovskiBatominovski

        1




        1























            -1












            $begingroup$

            Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is



            $$ frac{distance }{v cos alpha} $$



            So $alpha$ should also be known.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The theta wasn't given.
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:05










            • $begingroup$
              If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:09












            • $begingroup$
              Ooh I get that. How did you derive that equations
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:13










            • $begingroup$
              Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:32
















            -1












            $begingroup$

            Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is



            $$ frac{distance }{v cos alpha} $$



            So $alpha$ should also be known.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The theta wasn't given.
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:05










            • $begingroup$
              If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:09












            • $begingroup$
              Ooh I get that. How did you derive that equations
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:13










            • $begingroup$
              Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:32














            -1












            -1








            -1





            $begingroup$

            Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is



            $$ frac{distance }{v cos alpha} $$



            So $alpha$ should also be known.






            share|cite|improve this answer









            $endgroup$



            Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is



            $$ frac{distance }{v cos alpha} $$



            So $alpha$ should also be known.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 18 '15 at 0:03









            NarasimhamNarasimham

            20.7k52158




            20.7k52158












            • $begingroup$
              The theta wasn't given.
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:05










            • $begingroup$
              If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:09












            • $begingroup$
              Ooh I get that. How did you derive that equations
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:13










            • $begingroup$
              Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:32


















            • $begingroup$
              The theta wasn't given.
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:05










            • $begingroup$
              If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:09












            • $begingroup$
              Ooh I get that. How did you derive that equations
              $endgroup$
              – Zayn Malek
              Oct 18 '15 at 0:13










            • $begingroup$
              Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
              $endgroup$
              – Narasimham
              Oct 18 '15 at 0:32
















            $begingroup$
            The theta wasn't given.
            $endgroup$
            – Zayn Malek
            Oct 18 '15 at 0:05




            $begingroup$
            The theta wasn't given.
            $endgroup$
            – Zayn Malek
            Oct 18 '15 at 0:05












            $begingroup$
            If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
            $endgroup$
            – Narasimham
            Oct 18 '15 at 0:09






            $begingroup$
            If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
            $endgroup$
            – Narasimham
            Oct 18 '15 at 0:09














            $begingroup$
            Ooh I get that. How did you derive that equations
            $endgroup$
            – Zayn Malek
            Oct 18 '15 at 0:13




            $begingroup$
            Ooh I get that. How did you derive that equations
            $endgroup$
            – Zayn Malek
            Oct 18 '15 at 0:13












            $begingroup$
            Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
            $endgroup$
            – Narasimham
            Oct 18 '15 at 0:32




            $begingroup$
            Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
            $endgroup$
            – Narasimham
            Oct 18 '15 at 0:32


















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