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Find angle at which the projectile hits

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A kid stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.6 m/s. The cliff is 48.8 m above the ocean. How long after being released does the stone strike the water? With what speed and angle of impact does it hit the water?


    2. Relevant equations
    y = V_oy * t + 1/2 * g * t²
    x = v_ox * t

    3. The attempt at a solution
    Well to find the time I just used the first equation...plugged in -48.8 m for y, initial velocity in the y direction is 0.....g is going to be -9.8 here...and finally t²

    So t² = -48.8 / -4.9 ...and t ≈3.2 seconds

    Now for the distance I used the second equation...plugged 16.6 in for V_ox and the 3.2 for t
    which came out to ≈ 52.4m

    But the angle part is confusing me...how do I figure out the angle? I know that i would normally use the y components and x components....but I dont have an angle because the stone was thrown horizontally.... am I supposed to use the cliff 48.8sin(90) for the y component (no x component there)....and use 16.6cos(0) for the x component? (no y component there)....So would the answer be inverse tangent (48.8/16.6) = ≈71°?
     
  2. jcsd
  3. Feb 22, 2013 #2
    Hint: 48.8 meters is a distance, 16.6 is a velocity, and you want the relevant x and y components to be in the same units. And it doesn't matter that the stone was thrown horizontally, you're concerned some that's happening at the time when it hits the water.
     
  4. Feb 22, 2013 #3
    Hmm. that actually confused me more haha. i get what you mean...the cliff height is m and velocity is m/s...So then I can't use the cliff...okay

    Well I can't only use the projectile component can I? because there is no y component of it...with sin of 0 it would be 0.....I know I have the time right because it checkes out with the height of the cliff...the distance I'm pretty sure I have right V_ox * t......Perhaps another hint? I feel like it's something so incredibly obvious I'm going to hate myself for not realizing it.
     
  5. Feb 22, 2013 #4
    Haha, alright. Think about finding a slope first rather than an angle first. What you want is [itex] \frac{dy}{dt} [/itex] and [itex]\frac{dx}{dt} [/itex] when it hits the water. The downward velocity and the horizontal velocity when it hits the water are components of its total velocity. Turn this into a triangle and then solve for the angle that way. And it always helps to draw a picture.
     
  6. Feb 22, 2013 #5
    I don't know but ignore my answer OP, and correct me if i got it wrong helpers.
    Vx = 16.6m/s

    Vy = Vinitial + gt
    Vy = 0 + gt = 9.81m/s(3.2s)

    tan(theta) = Vy/Vx = 31.392/16.6
    theta = 62.13

    Vresult^2 = 31.392^2 + 16.6^2
    Vresult = 35.51
     
  7. Feb 23, 2013 #6
    Well what I was thinking was...using the first equation again

    y = V_oy * t + 1/2 * g * t²

    and assume I don't know the time yet

    48.8 = 0 + 1/2 (9.8) (t)²

    t = √((48.8/9.8)/2)

    which WOULD comfirm my time of ≈3.2 seconds

    Now since thats ONLY in the y direction....shouldn't the velocity = acceleration * time?

    So 9.8 ( 3.2 ) = 31.36 would be my Vy

    so now

    tanθ Vy/Vx = 62.11°

    and the speed would be √(31.36² + 16.6²)

    which comes out to 35.1 m/s


    Our answers are a bit off from each other...anyone out there that can tell us who was in the right direction?
     
  8. Feb 23, 2013 #7
    With the precision both computations are done, the last digit BEFORE the decimal is already off in both computations, so it is meaningless to compare the less significant digits. If that sort of precision is OK, then both computations are correct.
     
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