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Monkey and Hunter:Projectile Motion

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the setup of a gun aimed at a target as shown in the figure. The target is to be dropped from point A at t = 0 s, the same moment as the gun is fired. The bullet hits the target at a point P which is along vertical line to the ground. Let the initial speed of the bullet be Vi= 149 m/s and the angle between the vector Vi and the x-axis be θ = 61.4 degrees and the starting height of the target be 87.3 m. The horizontal distance lies on the x-axis.
    The acceleration of gravity is 9.8 m/s2.

    a) Determine the y component of the velocity of the bullet at the time of collision.
    Answer in units of m/s

    b) At what height, above the ground, do the bullet and target hit.
    Answer in units of m

    c) Find the speed of the target at the point of collision.
    Answer in units of m/s

    2. Relevant equations

    -None-

    3. The attempt at a solution

    I believe I am correct for these answers, but when I enter them into a website where we submit our work, it says the answers are wrong for everything (a-c). Am I doing something wrong?

    a) Vyimpact=Vyi+at
    Delta x = 87.3tan(61.4)
    From that: 160.1195=149cos(61.4)t
    t=2.2449

    Vyimpact=Vyi+at
    Vyimpact= 149sin(61.4)-(9.8)(2.2449)=108.819 m/s

    b) Distance target fell in 2.2449 seconds= -(1/2)(a)t2
    delta y= -(.5)(9.8)(2.2449)2 = -24.6939

    Distance target was hit above ground = 87.3-24.6939=62.6061 m

    c) Vxi=Vx2 *Neglecting any horizontal factors
    Vx2=149cos(61.4)=71.3251 m/s
    Vy2=Vyimpact
    Vy2= 108.819

    Vimpact= sqrt(71.32512 + 108.8192)= 130.1108 m/s
     
  2. jcsd
  3. Oct 11, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello and welcome to PF!

    Check to see if you used the trig function correctly here.
    Otherwise, your work looks good to me. :)
     
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