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Projectile from certain height

  1. Apr 16, 2015 #1
    Formula for time of flight in projectile motion is 2usin theta/g but I think it is only applicable for when object is launched from ground i.e in the case below
    Projectile-motion-animation.png
    But when it is launched from certain height this formula is no longer in use.
    images?q=tbn:ANd9GcRjK4KyeUxJLwX90VQcMPICChwZ8V-Z5iL49ZLZ4wSKeiaxe-sE6Q.jpg
     
    Last edited: Apr 16, 2015
  2. jcsd
  3. Apr 16, 2015 #2

    ZapperZ

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    Is there a question here?

    Zz.
     
  4. Apr 16, 2015 #3
    Am I right in my first post.
     
  5. Apr 16, 2015 #4
    Yes, it does not work in the second case.
     
  6. Apr 16, 2015 #5
    Thanks @nasu.
     
  7. Apr 16, 2015 #6
    And what about range and maximum height.Formula for range=u^2sin 2 theta/g
    and maximum height =u^2 sin^2 theta/2g
    Are these formulas also not applicable for second case.
     
  8. Apr 16, 2015 #7
    I think it would be more beneficial for you to try to answer these starting from the basic equations.
    This is the way to study. Not memorizing some formulas for special cases.
     
  9. Apr 17, 2015 #8
    I think ,you are right.I have done some calculations.Check whether I am correct.
     
  10. Apr 17, 2015 #9
    I have divided the trajectory in to two parts.
    upload_2015-4-17_12-20-43.png

    Now
    S sub y (distance covered in y direction)=V0 sin theta T +1/2 g T^2
    h=V0 sin theta T+1/2 g T^2
    where T =time of flight
    Now we can solve for T as other variables are given.The above equation forms quadratic equation.so we will gwt two values for T one of which will be negative ,as time can never be negative we will take the other one(positive value of T)
    (I have taken downward direction to be positive)
    And S sub x=(distance covered in x direction)=V0 cos theta t
    If we put t=T in the equation we get
    Vo cos theta T
    which is nothing but Range.
    For maximum height
    I have again divided the trajectory in two parts

    upload_2015-4-17_12-37-0.png
    Now we can see the maximum height=h+h'
    where h' is same as in case of object projected from horizontal surface and lands on the same horizontal surface.And h is given,hence we can find maximum height.
    Is my work fine?
     
  11. Apr 17, 2015 #10

    jbriggs444

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    If you take the downward direction to be positive then your formula for Sy as a function of time has the wrong sign on the contribution from the object's initial velocity.
     
  12. Apr 17, 2015 #11
    It should be as follows.
    h= - V0 sin theta T +1/2g T^2
    Right?
     
  13. Apr 17, 2015 #12

    jbriggs444

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  14. Apr 17, 2015 #13
    Thanks.
     
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