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Homework Help: Projectile motion confusion!

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A stone is thrown horizontally from a height of 10m above ground and moves with constant velocity of 5m/s. Calculate:

    a) the time it will take to hit the ground
    b) the distance traveled in the calculated time


    2. Relevant equations

    Now for a i am using x = vot+1/2at2 but i just cannot get my head around it
    I understand that b will be velocity x time?


    3. The attempt at a solution

    10 = 0 x t +1/2 x 9.81 xt2
    10 = ??????

    I just have some kind of mental block going on, i've read my text books and websites but no luck.
     
  2. jcsd
  3. Jan 17, 2010 #2
    You have left out the term which involves initial position but since your acceleration is positive you will still get the right answer. Other than that, you're incredibly close. Try using Latex in the future so we can read your equations without getting a headache.
     
  4. Jan 17, 2010 #3
    first off:
    g = 9.81

    y = y0 + v0y*t - (1/2)*g* t ^2

    (notice the minus, since the accelaration is negative - gravity pulls the body down, if we convention the y axis is pointing up)
    also, notice y0 (inital position in the vertical axis) = 10
    and the vertical component of the initial velocity = 0
    so:
    1) y = 10 - 1/2*g*t^2

    also : x = x0 + v0x*t +1/2 a t^2
    so
    we know that the horizontal accelaration, and the horizontal position at t=0 equal 0, and v0 = 5 so:
    2) x = 0 + 5*t + 0

    We want to know at what time the object hits the ground, so y = 0. substitute it in the 1) equation.

    Solve the equation:

    0 = 10 - 1/2*9.8*t^2.

    then, substitute t in the 2) equation and solve it.
     
  5. Jan 17, 2010 #4
    I still don't get it :/ This is driving me insane. My physics teacher is useless, we never do enough examples.
     
  6. Jan 17, 2010 #5
    okay:
    First of all, i recomend you try and draw the problem, as it always help to understand.
    So, you need two equations, one for the horizontal movement, and one for the vertical, since the object will move both horizontally (it has a horizontal speed), and vertically (it's falling)

    it's initial position can be given by the vector (x,y) = (0,5)

    Horizontal :
    [tex]x = x_{0} + v_{x} \times t + \frac{1}{2} \times a \times t^{2}[/tex]

    Vertical:
    [tex]y = y_{0} + v_{y} \times t + \frac{1}{2} \times g \times t^{2}[/tex]

    g = 9.81

    first you want to know at what time it hits the ground, so, you just need the vertical equation. It hits the ground when y = 0.
    [tex]y_{0}[/tex] , is the initial vertical position. Since it's 10m above the ground, it equals 10.
    v_{y} = 0, because there is no initial velocity downwards or upwards, and the accelaration is, as stated, 9.81. Since the movement is downwards, the accelaration is negative (of course this depends on the referential you choose, but generally speaking, it's simpler this way)
    Therefor :

    Vertical:
    [tex]0 = 10 + 0 \times t - \frac{1}{2} \times g \times t^{2}[/tex]

    [tex]t = \sqrt{\frac{10 \times 2}{g}}[/tex]

    solve that, and you have the time it hits the floor.

    Then you need to know the distance it traveled.

    Now, you need the horizontal equation.
    We know it's initial horizontal position is 0, initial velocity 5, and there is no accelaration given, so:

    [tex]x = 0 + 5 \times t + 0[/tex]

    Now, take the t you got from your first equation, and substitute it here. Now you do the math.

    This is the general basis for projectile based problems: You should remember it.
    You have to consider always an equation for each axis of movement, consider the velocity in each axis, the initial position in each axis, and the accelaration.
     
  7. Jan 17, 2010 #6
    Thankyou Fanta, makes sense now.
     
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