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Projectile Motion of a Shot Problem

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A projectile is shot from the edge of a cliff 115 above ground level with an initial speed of 65.0 at an angle of 35.0 with the horizontal, as shown in the figure .

    At the instant just before the projectile hits point P, determine the horizontal and the vertical components of its velocity.
    Enter your answers numerically separated by a comma.

    2. Relevant equations

    basic kinematic equations [a= constant]

    3. The attempt at a solution

    My attempt was pretty basic, I thought it was correct, and even asked my brother whos a physics tutor for help and he got the same answer, though the MasteringPhysics site which my class is using says its wrong

    the horizontal x direction I got the same speed as it is when shot from the start so it doesn't change

    65m/s*cos(35o) = 53.2 m/s in the horizontal direction

    for the vertical I used the equation

    v = vo + at

    where vo = 65m/s*sin(35o) = 37.3 m/s
    a = 9.8 m/s2
    t = 2.4 s

    "t" I found in a different part of the question.

    so "v" = 60.36 m/s

    though the site says its wrong (masteringphysics)


    2. Relevant equations
  2. jcsd
  3. May 23, 2010 #2
    Use the formula V = (U^2 + 2gh)^1/2
    to find the final velocity
  4. May 23, 2010 #3
    I dont think I've seen that formula before, where you get it? and U is velocity initial right?

    also its still equals 60.38, which I already used as a former answer, does that mean my horizontal answer is wrong or the teacher who posted this question online is wrong?
  5. May 23, 2010 #4
    hmm.. It looks like your teacher may have made a mistake then.
    what was his answer?

    and yes "U" is initial Velocity
  6. May 23, 2010 #5
    there is no answer, its a site called mastering physics, he posts questions and we input an answer if its right we get credit if wrong we lose.
  7. May 24, 2010 #6


    User Avatar
    Homework Helper

    Can you show your calculation of t?
    I am using the formula

    -115 = 37.3(t) - 0.5*9.8(t)^2.

    Solve the quadratic to find t.
  8. May 24, 2010 #7
    115 = 0 + (3.73m/s)*t + 4.9t2

    since velocity and acceleration are moving in the same direction at this point

    then using the quadratic equation

    -37.3 (+/-)sqr(37.32 - 4(4.9)(-115))/ 2*4.9

    which ended up 2.355102041 and some other number I didn't record.
  9. May 24, 2010 #8


    User Avatar
    Homework Helper

    115 = 0 + (3.73m/s)*t + 4.9t^2

    The displacement and the acceleration are in the same direction.

    But the velocity is in the opposite direction to acceleration. So the equation should be

    115 = 0 - (3.73m/s)*t + 4.9t2
  10. May 24, 2010 #9
    ah, I got a email back from my teacher

    since its going downwards the velocity component is negative. woops
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