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Homework Help: Physics - Projectile Motion

  1. Jul 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown with a velocity of 20.0 m/s [30 degrees] and travels for 3.0 s before it strikes the ground. Find the
    a) height from which it was thrown.
    b) the maximum height of the ball.

    2. Relevant equations
    a) dx = vx • t ... I don't even know what to do for part A.
    b) v2f = v2i + 2ad

    3. The attempt at a solution
    a) I absolutely have no clue what to do for part A.

    I first thought to use triangles and tried to find the distance the ball travelled using the velocity I found, but it doesn't even work.

    b) I realized that at the maximum height, the velocity of the object will me 0 m/s, so I tried to find the distance from that point to the ground to find the maximum height.

    I found the initial velocity to be 10 m/s using SOH and i used the above formula.
    viy: 10 m/s
    vfy: 0 m/s
    g(a): -9.81 m/s2

    02 m/s= 102 m/s + 2 • -9.81 m/s2 • d
    0 m/s = 100 m/s + -19.62 m/s2 • d
    -100 m/s = -19.62 m/s2 • d
    d = 5.1 m

    It's not the right answer, so I don't know where I went wrong.

    I don't know if any of what I did makes sense, I'm so lost in projectile motion.
  2. jcsd
  3. Jul 8, 2010 #2


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    Homework Helper

    Hi rissie, welcome to PF.

    When the ball reaches the ground, the displacement of the ball is d, the height from which the ball is thrown. Initial velocity is 10 m/s. The total time is 3 s.
    Use the formula
    -d = V(iy)*t - 1/2*g*t^2.

    Substitute the values and find d.
  4. Jul 8, 2010 #3
    Ohhh, okay. That makes so much more sense now.

    That's just for the height it was launched from, right?
  5. Jul 9, 2010 #4


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    Homework Helper

    Yes. For maximum height add 5.1 m to d.
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