1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Problem on projectile

  1. Jul 10, 2013 #1
    See the attachment.

    The right answer is θ=60 degrees

    but i am getting sinθ = ±√3.

    Attached Files:

  2. jcsd
  3. Jul 10, 2013 #2
    Your expression for ##v_y## is wrong.

    Also, check your last step when you plug 3 in 4.
    Last edited: Jul 10, 2013
  4. Jul 10, 2013 #3


    User Avatar
    Gold Member

    I don't agree with your very first equation. It is not true that
    [tex]H= \frac{u^{2} \sin^{2}( \theta)}{2g}.[/tex]
    Conservation of Energy requires, instead, that
    [tex]mgH+ \frac{m u^{2} \cos^{2}( \theta)}{2}= \frac{mu^{2}}{2},[/tex]
    comparing the peak to the starting-point, or
    [tex]gH+ \frac{u^{2} \cos^{2}( \theta)}{2}= \frac{u^{2}}{2}.[/tex]
    Then you can also write
    [tex] \frac{gH}{2}+ \frac{5 u^{2} \cos^{2}( \theta)}{4}= \frac{u^{2}}{2},[/tex]
    comparing the energies at the half-way point to the starting-point.
  5. Jul 10, 2013 #4
    First things first.

    How is my expression for vy wrong?

    I used v2-u2 = 2as
  6. Jul 10, 2013 #5
    What are the directions for velocity in vertical direction and acceleration?
  7. Jul 10, 2013 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Doesn't that reduce to the same equation?
  8. Jul 10, 2013 #7


    User Avatar
    Gold Member

    You're quite right. Chalk that one up to needing to do one step at a time. Also, for the half-way point, it doesn't reduce the same way.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted