Angry Birds Question in Physics Exam
By on July 10th, 2011

We have all come to love and have been playing the game on , , PSP, Desktop, Chrome and most recently Windows Phone 7 devices. In fact, we are also so crazy about it that we have been buying some cool Angry Birds merchandise as well.

As you can see, those tiny little birds have taken the world by storm and are in every nook or corner possible. In fact, there are even movies being made on them. So will this be the end? Apparently not. Some school decided to actually use Angry Birds as a question in their physics paper.

The question was no joke and could fetch students 7 marks. This is definitely something. Incidentally, there have also been other instances where teachers have used Angry Birds to teach physics to students.


So  what will these Angry Birds invade next?

(via Reddit)

Tags: ,
Author: Keith Dsouza Google Profile for Keith Dsouza
I am the editor-in-chief and owner of Techie Buzz. I love coding and have contributed to several open source projects in the past. You can know more about me and my projects by visiting my Personal Website. I am also a social networking enthusiast and can be found active on twitter, you can follow Keith on twitter @keithdsouza. You can click on my name to visit my Google+ profile.

Keith Dsouza has written and can be contacted at
  • Karthik Prabhu

    Now that was something really interesting. :D People have really become addicted to this game. Simple concept, but a very addictive game.

  • Meena Bassem

    quite nice to have up-to-date teachers :D
    btw, if i had a question like would be in the maths(dynamics) test , not the physics one.
    but really like the idea quite funny and not that difficult to solve

  • Eric Garza

    The sketched solution will slightly miss the target, the pig is 2m above the bird but the student is assuming they are at the same level.

  • Sanjevi

    This problem can be solved by using the theory of ‘projectile motion’, the formula is

    time(t) = velocity(v) x sin(angle)/g
    i.e. sin(angle) = v / t x g
    sin(angle)= 55.0 / 2.50 x 9.8
    sin(angle) = 0.8980
    angle = 63degree 54min ………….ans.

  • Deep Singh Nanuan

    Horizontally :
    x = 55.0m
    Vix = 22.0 m/s
    ax = 0 m/s/s

    Vertically :
    y = (12-10) = 2 m
    Viy = 0 m/s = 22tanÆŸ
    ay = -9.8 m/s/s

    ÆŸ = (tan^-1)(32.625/55)
    ÆŸ = 30.68 degrees

  • Adel Ansari

    My answer :D

    x= 55m , y= 2m and t= 2.50 s.
    Vx= V.Cos(angle)
    Vx=x / t => Vx=55/2.50 = 22 m.s^-1.
    y= (-1/2) × g t^2 + V t Because (angle).
    2=(-1/2) × 9.8 × 2.50^2 + 2.50 × ( Vx / Cos( angle) )× Sin( angle).
    2+30.625= 2.50 × 22 × tan( angle).
    tan( angle) = 32.625 / 55.
    tan( angle) = 0.59318.
    angle = tan^-1 ( 0.59318).
    angle = 30.67564743.

  • Adel Ansari

    My answer :D

    x= 55m , y= 2m and t= 2.50 s
    Vx= V.Cos(angle)
    Vx=x / t => Vx=55/2.50 = 22 m.s^-1
    y= (-1/2) × g t^2 + V t Cos (angle)
    2=(-1/2) × 9.8 × 2.50^2 + 2.50 × ( Vx / Cos( angle) )× Sin( angle)
    2+30.625= 2.50 × 22 × tan( angle)
    tan( angle) = 32.625 / 55
    tan( angle) = 0.59318
    angle = tan^-1 ( 0.59318)
    angle = 30.67564743

  • Pingback: Batman Equation - Epic Fun

Copyright 2006-2012 Techie Buzz. All Rights Reserved. Our content may not be reproduced on other websites. Content Delivery by MaxCDN