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Finding Range, Max Height and Speed of Projectile

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired with an initial speed of 100 ft/s and angle of elevation Pi/6 (30 degrees).
    A) the range of the projectile (along the x-axis)
    B) the maximum height reached
    C) the speed at impact

    2. Relevant equations

    Vf^2 = Vi^2 +2a*s
    Vf = Vi + a*t
    Sf = Si + Vo*t + (1/2)a*t^2
    (where S is any coordinate axis)

    3. The attempt at a solution
    I have the answers from the answer key (this is a review), and I can't seem to get any of it right.
    I found s (max height) (not S), to be 127.551 seconds.
    I found t = 10.2 (from launch to landing)
    I found range = 883.346.

    I know this isn't a hard problem, but for some reason I cannot solve it.
    The answers are:
    A) 625*sqrt(3)/4 ft
    B) 625/16 ft
    C) 100 ft/s
  2. jcsd
  3. Dec 12, 2008 #2
    How did you obtain those answers?

    Try separating the equations for position into the x and y components.

  4. Dec 12, 2008 #3
    Vf = 0 (on the way up)
    0 = 100^2*sin(30)^2-2(9.8)s
    solving for s gives = 127.551 ft.

    for time on the way up:
    127.551 = 0 + 100sin(30)t - 4.8t^2
    solving for t gives = 5.1 seconds

    xf = 0 + 100cos(30)(5.1)2
    = 883.346

    I didn't try finding the impact speed because the other two answers were wrong.
  5. Dec 12, 2008 #4


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    Homework Helper
    Gold Member

    The problem velocity is given in units of feet per second, not meters per second. Check out your value of 'g'.
  6. Dec 27, 2008 #5
    There's no acceleration along x-axis. The acceleration due to gravity acts only along the -y axis. 'A' implies the angle of projection.
    Assuming the projectile is fired from the origin,
    The net y-displacement of the projectile is zero, since the projectile returns on x-axis.
    Substituting t,
    This is the general formula for range of projectile.
    Using given data,
    x=269.20 ft ...(Range)

    At the max height, the y-velocity of projectile is zero.
    This is the general forumla for max height of projectile.
    Using given data,
    y=38.86 ft ...(Max height)

    Since, there is no acceleration along x-axis,
    The net y-displacement of projectile is zero
    The components of final velocity are same as that of initial velocity, hence, the velocities must be equal. Therefore,
    v=100 ft/s
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