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Projectile Motion Calculation Question

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A projectile has a time of flight of 7.5s and a range of 1200m. Calculate:
    a) Its horizontal velocity (This one was done correctly, just posting for follow through reference)

    b) Its maximum height

    c) The velocity with which it is projected


    2. Relevant equations
    [tex]\Delta[/tex]x=uxt

    [tex]\Delta[/tex]y=uyt + 1/2 * ayt2

    uy=usin[tex]\theta[/tex]

    3. The attempt at a solution

    a)
    1200=ux*7.5
    ux=160ms-1 (This is correct)

    b)
    7.5 = (2usin[tex]\theta[/tex])/9.8
    2usin[tex]\theta[/tex]=73.5 ---- Equation 1

    ux=ucos[tex]\theta[/tex]
    160=ucos[tex]\theta[/tex]
    u = 160/cos[tex]\theta[/tex] ---- Equation 2

    Sub Equation 2 into 1:
    2(160/cos[tex]\theta[/tex])*sin[tex]\theta[/tex]=73.5
    320tan[tex]\theta[/tex]=73.5
    tan[tex]\theta[/tex]=73.5/320
    [tex]\theta[/tex]=12degree 56 min (nearest min)

    u=164.17ms-1

    uy=usin[tex]\theta[/tex]
    =164.17*sin12deg56min
    =36.75ms-1

    [tex]\Delta[/tex]y=(36.75*7.5)+(0.5*9.8*7.52)

    [tex]\theta[/tex]y=551.25m (This is my final answer but it isn't the answer in the book :()

    c) Yeah u=164.17ms-1 But I can't get direction. I stated that direction was 12deg56min from ground, but it's wrong...

    Thanks alot if you can help me

    Cheers
     
  2. jcsd
  3. Sep 29, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    To calculate maximum height , you have to modify the equation LaTeX Code: \\Delta y=uyt + 1/2 * ayt2 by LaTeX Code: \\Delta y=uyt - 1/2 * ayt2. And for time you have to take half the time of flight.
     
  4. Sep 29, 2008 #3
    Hi,

    Thanks for help but why do I need to make it -1/2 ayt2?

    Thanks
     
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