Dee asked in Science & MathematicsMathematics · 2 months ago

physic questions p2?

Exam coming up, prof did not provide solutions. Anything helps, prefer if you show approach.

Thanks, 

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    I'm going to use a bit of calculus for my own clarification.  No need to pay attention just yet

    a(t) = -9.8

    v(t) = -9.8 * t + C

    s(t) = -4.9 * t^2 + Ct + D

    s(0) = 50.0

    s(t) = 50 + C * t - 4.9 * t^2

    Now you can pay attention.

    The first stone is thrown downward with a velocity of 2.00 m/s, so C = -2 (it's negative because velocity is a vector and we're moving downward)

    Let m(t) be the function for the first stone

    m(t) = 50 - 2t - 4.9t^2

    Find when m(t) = 0

    0 = 50 - 2t - 4.9 * t^2

    4.9 * t^2 + 2t - 50 = 0

    49t^2 + 20t - 500 = 0

    t = (-20 +/- sqrt(400 + 4 * 49 * 500)) / 98

    t = (-20 +/- 2 * sqrt(100 + 49 * 500)) / 98

    t = (-20 +/- 20 * sqrt(1 + 49 * 5)) / 98

    t = -20 * (1 +/- sqrt(1 + 245)) / 98

    t = -10 * (1 +/- sqrt(246)) / 49

    It wouldn't make sense for t < 0

    t = -10 * (1 - sqrt(246)) / 49

    t = 10 * (sqrt(246) - 1) / 49

    t = ‭2.9968137023179840682009503010742‬

    We can say that t = 3.00

    n(t) will be our next stone's path

    n(t) = -4.9 * t^2 + Ct + 50

    When t = 2, since 2 = 3 - 1, then n(t) = 0

    0 = -4.9 * 2^2 + C * 2 + 50

    4.9 * 4 - 50 = 2C

    4.9 * 2 - 25 = C

    9.8 - 25 = C

    -15.2 = C

    It's thrown down at 15.2 m/s

    m'(t) = -2 - 9.8 * t

    m'(3) = -2 - 9.8 * 3 = -2 - 29.4 = -31.4

    n'(t) = -15.2 - 9.8 * t

    n'(2) = -15.2 - 19.6 = -34.8

    Look, I'm sure that the professor didn't give you solutions, but it's not like this is the first time you've seen problems like these.  Draw some diagrams, look at your notes for projectile motion, and work it out, step by step.  Sometimes, that's all you can do.

    150 * cos(120) + m * cos(t) = 140 * cos(35)

    150 * sin(120) + m * sin(t) = 140 * sin(35)

    m * cos(t) = 10 * (14 * cos(35) - 15 * cos(120))

    m * sin(t) = 10 * (14 * sin(35) - 15 * sin(120))

    m^2 * (cos(t)^2 + sin(t)^2) = 10^2 * (196 * cos(35)^2 - 420 * cos(35) * cos(120) + 225 * cos(120)^2 + 196 * sin(35)^2 - 420 * sin(35) * sin(120) + 225 * sin(120)^2)

    m^2 * 1 = 10^2 * (196 + 225 - 420 * cos(120 - 35))

    m^2 = 100 * (421 - 420 * cos(85))

    m = 10 * sqrt(421 - 420 * cos(85))

    m = ‭196.05983475612325607405880556032‬

    150 * cos(120) + 196 * cos(t) = 140 * cos(35)

    196 * cos(t) = 140 * cos(35) - 150 * cos(120)

    cos(t) = (140 * cos(35) - 150 * cos(120)) / 196

    t = cos-1((140 * cos(35) - 150 * cos(120)) / 196)

    t = +/- ‭14.588063297951140303009775328656‬

    150 * sin(120) + 196 * sin(t) = 140 * sin(35)

    sin(t) = (140 * sin(35) - 150 * sin(120)) / 196

    sin(t) = (70 * sin(35) - 75 * sin(120)) / 98

    t = sin-1((70 * sin(35) - 75 * sin(120)) / 98)

    t = ‭-14.659673478957222794754062525788‬

    The discrepancy comes from my rounding of the magnitude before.  The direction is at -14.6 degrees with respect to the x-axis.

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  • Anonymous
    2 months ago

    Nah, do your own homework

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