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Math Help - Story drop Problems

  1. #1
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    Story drop Problems

    How do I solve these?

    A stone is dropped from the edge of a roof, and hits the ground with a velocity of -165 feet per second. How high (in feet) is the roof?
    and

    A particle is moving with acceleration a(t) = 24t+4. Its position at time t=0 is s(0) = 14 and its velocity at time t=0 is v(0) = 3 What is its position at time t=6? ________
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by killasnake View Post
    A stone is dropped from the edge of a roof, and hits the ground with a velocity of -165 feet per second. How high (in feet) is the roof?
    v_i=0 because it was from rest

    v_f=-165 I suppose down is negative in this problem

    a=-9.8 If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)

    \Delta d=? This is what we're trying to find

    Notice that they don't give time, so we'll use an equation without time.

    The only one that comes to mind is: v_f^2=v_i^2+2a\Delta d

    So then we switch things around to get: \Delta d=\frac{v_f^2-v_i^2}{2a}

    So substitute to get the answer
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  3. #3
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    Quote Originally Posted by Quick View Post
    v_i=0 because it was from rest

    v_f=-165 I suppose down is negative in this problem

    a=-9.8 If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)

    \Delta d=? This is what we're trying to find

    Notice that they don't give time, so we'll use an equation without time.

    The only one that comes to mind is: v_f^2=v_i^2+2a\Delta d

    So then we switch things around to get: \Delta d=\frac{v_f^2-v_i^2}{2a}

    So substitute to get the answer
    That does fairly well for an answer Quick, except for one little problem:
    Acceleration is in units of m/s^2, not m/s. (Or ft/s^2 if you prefer.)

    Tsk, tsk! This is the second time I've had to tell you that!

    -Dan
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    Quote Originally Posted by Quick View Post

    a=-9.8 If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)
    s(t) = -16t^2 ft
    v(t) = -32t ft/s
    a(t) = -32 ft/s^2
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by killasnake View Post
    A particle is moving with acceleration a(t) = 24t+4. Its position at time t=0 is s(0) = 14 and its velocity at time t=0 is v(0) = 3 What is its position at time t=6?
    We've got the acceleration function, and we know s(0) and v(0) as initial conditions. So:
    v(t) - v(0) = \int_0^t \, a(t') \, dt' = \int_0^t \, (24t' + 4)dt' = (12t'^2 + 4t')|_0^t = 12t^2 + 4t

    So v(t) = 12t^2 + 4t - 3

    Again:
    s(t) - s(0) = \int_0^t \, v(t') \, dt' = \int_0^t \, (12t'^2 + 4t' - 3)dt' = (4t'^3 + 2t'^2 - 3t')|_0^t = 4t^3 + 2t^2 - 3t

    So s(t) = 4t^3 + 2t^2 - 3t - 14

    Thus s(6) = 4(6)^3 + 2(6)^2 - 3(6) - 14 = 904 (in whatever units the problem is supposed to be written in.)

    Piece of trivia for you: The coefficient of the t^3 term is constant. This means that the above equation for displacement is written in terms of a constant "jerk." (No kidding! The jerk, j, is defined as the first time derivative of the acceleration.)

    -Dan
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  6. #6
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    wow you guys are quick! thank you so much for the help!
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  7. #7
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    Hello, killasnake!

    1) A stone is dropped from the edge of a roof,
    and hits the ground with a velocity of -165 ft/sec.
    How high (in feet) is the roof?

    You're expected to know the "free fall" formula: . y(t) \:=\:h_o + v_ot - 16t^2

    . . where: . \begin{Bmatrix}y(t) = & \text{height at time }t \\ h_o = & \text{initial height} \\ v_o = & \text{initial velocity} \\ t = & \text{time in seconds}\end{Bmatrix}

    In this problem, the stone is dropped: v_o = 0
    . . Hence, the function is: . y(t) \:=\:h_o - 16t^2
    and we want to determine h_o.

    We are told that the stone hits the ground at -165 ft/sec.
    When does this happen?
    . . It happens when y(t) = 0
    So we have: . h_o - 16t^2\:=\:0\quad\Rightarrow\quad t = \frac{\sqrt{h_o}}{4} seconds.

    Velocity is the derivative of position: . v(t)\:=\:-32t

    At that time, we have: . -32t \:=\:-165\quad\Rightarrow\quad-32\left(\frac{\sqrt{h_o}}{4}\right)\:=\:-165

    . . -8\sqrt{h_o} \:=\:-165\quad\Rightarrow\quad \sqrt{h_o}\:=\:\frac{165}{8}\quad\Rightarrow\quad h_o \:=\:\left(\frac{165}{8}\right)^2

    Therefore: . h_o\:=\:425.390625 \:\approx\:425.4 feet.

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  8. #8
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    Quote Originally Posted by topsquark View Post
    We've got the acceleration function, and we know s(0) and v(0) as initial conditions. So:
    v(t) - v(0) = \int_0^t \, a(t') \, dt' = \int_0^t \, (24t' + 4)dt' = (12t'^2 + 4t')|_0^t = 12t^2 + 4t

    So v(t) = 12t^2 + 4t - 3

    Again:
    s(t) - s(0) = \int_0^t \, v(t') \, dt' = \int_0^t \, (12t'^2 + 4t' - 3)dt' = (4t'^3 + 2t'^2 - 3t')|_0^t = 4t^3 + 2t^2 - 3t

    So s(t) = 4t^3 + 2t^2 - 3t - 14

    Thus s(6) = 4(6)^3 + 2(6)^2 - 3(6) - 14 = 904 (in whatever units the problem is supposed to be written in.)
    -Dan
    Slight correction on this problem.

    v(t) is NOT v(t) = 12t^2 + 4t - 3
    it is v(t) = 12t^2 + 4t + 3

    and hence

    s(t) is NOT s(t) = 4t^3 + 2t^2 - 3t - 14
    it is s(t) = 4t^3 + 2t^2 + 3t + 14

    so the answer is 968 not 904
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