At a basketball game, members of the spirit club use a t-shirt cannon to launch team t-shirts to spectators in the stands.

At a basketball game, members of the spirit club use a t-shirt cannon to launch team t-shirts to spectators in the stands. the equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v. h = -16t 2 + vt + c solve the equation for initial velocity, v. the cannon launched a t-shirt from an initial height of 5 feet. after 3 seconds the shirt has reached a height of 131 feet. use your equation from #1 to find the initial velocity of the t-shirt. show your work. what does the equation from #1 tell you about the value of the velocity, v, when t = 0? explain.

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  1. The equation for initial velocity is s = ut + 1/2at^2, where s = distance, u = initial velocity, and a = acceleration.

    Step-by-step explanation:

  2. The two solutions are given as [tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex] and [tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]

    Step-by-step explanation:

    As the given equation is

    [tex]y''+7y'-8y=0\\[/tex]

    So the corresponding equation is given as

    [tex]m^2+7m-8=0[/tex]

    Solving this equation yields the value of m as

    [tex](m+8)(m-1)=0\\m=-8, m=1[/tex]

    Now the equation is given as

    [tex]y(t)=C_1e^{m_1t}+C_2e^{m_2t}[/tex]

    Here m1=-8, m2=1 so

    [tex]y(t)=C_1e^{-8t}+C_2e^{t}[/tex]

    The derivative is given as

    [tex]y'(t)=-8C_1e^{-8t}+C_2e^{t}[/tex]

    Now for the first case y(t=0)=1, y'(t=0)=0

    [tex]y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2[/tex]

    So the two equation of co-efficient are given as

    [tex]C_1+C_2=1\\-8C_1+C_2=0[/tex]

    Solving the equation yield

    [tex]C_1=1/9 \\C_2=8/9[/tex]

    So the function is given as

    [tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex]

    Now for the second case y(t=0)=0, y'(t=0)=1

    [tex]y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2[/tex]

    So the two equation of co-efficient are given as

    [tex]C_1+C_2=0\\-8C_1+C_2=1[/tex]

    Solving the equation yield

    [tex]C_1=-1/9 \\C_2=1/9[/tex]

    So the function is given as

    [tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]

    So the two solutions are given as [tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex] and [tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]

  3. 1. h = -16t² + vt + c
    ⇒ vt = 16t² + h - c
    ⇒ v = 16t + (h-c)/t

    2. c = 5 ft
    t = 3s
    h = 131 ft

    v = 16*3 + (131-5)/3 = 48 + 42 = 90 ft/s

    when t=0, velocity is not defined. That means at t=0, there is no velocity of the t-shirt or it is 0.

  4. a. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]

    b. [tex]C = 950[/tex]

    c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]

    Step-by-step explanation:

    a. Let the number of penguins who have the disease t days after the outbreak be P

    Initial number of penguins = 1000

    Therefore, current number of penguins = 1000 - P

    And the rate of spread of disease according to the statement is

    [tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]

    where k is the constant of proportionality

    [tex]\frac{dP}{1000-P}=kt.dt[/tex]

    Integrating both sides

    [tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]

    b. Seeing as 50 penguins had the disease initially,

    t = 0

    P = 50

    The general solution of the differential solution becomes

    50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)

    [tex]C = 1000 - 50 = 950[/tex]

    c. Therefore, the solution that satisfies the initial condition is

    [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]

  5. A) Judging by the slope field, it would appear the solution passing through (0, 2) would just be the line [tex]y=2[/tex]. This holds up for the given ODE, since if [tex]y(x)=2[/tex], then both sides of the ODE reduce to 0.

    Since we can surmise that [tex]y=2[/tex] is an equilibrium for this ODE (that is, the derivative along this line is 0 everywhere), and the slope at (1, 0) is positive, we know that as [tex]x\to+\infty[/tex], the function [tex]y(x)[/tex] will converge to 2. In other words, as [tex]x[/tex] gets larger, the slope field suggests that the solution curve through (1, 0) will start to plateau and steadily approach [tex]y=2[/tex]. On the other hand, as [tex]x\to-\infty[/tex], the slope field tells us that the curve would rapidly diverge off to [tex]-\infty[/tex]. (When you actually draw the solution, you would end up with something resembling the plot of [tex]-e^{-x}[/tex].)

    b) The tangent line to [tex]y=f(x)[/tex] at [tex]x=1[/tex], given that [tex]f(1)=0[/tex], takes the form

    [tex]\ell(x)=f(1)+f'(1)(x-1)[/tex]
    [tex]\ell(x)=f'(1)(x-1)[/tex]

    When [tex]x=1[/tex], we have [tex]y=0[/tex], so

    [tex]f'(1)=\dfrac13\cdot1\cdot(0-2)^2=\dfrac43[/tex]

    and so the tangent line to [tex]f(x)[/tex] at [tex]x=1[/tex] is

    [tex]\ell(x)=\dfrac43(x-1)[/tex]

    Using the tangent line as an approximation, we would find

    [tex]f(0.7)\approx\ell(0.7)=\dfrac43\left(\dfrac7{10}-1\right)=-\dfrac4{10}=-0.4[/tex]

    c) The ODE is separable, so we can write

    [tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac13x(y-2)^2\iff\dfrac{\mathrm dy}{(y-2)^2}=\dfrac x3\,\mathrm dx[/tex]

    Integrating both sides gives us

    [tex]-\dfrac1{y-2}=\dfrac{x^2}6+C[/tex]

    Given that [tex]y(1)=0[/tex], we get

    [tex]-\dfrac1{0-2}=\dfrac{1^2}6+C\implies C=\dfrac13[/tex]

    so the particular solution is

    [tex]-\dfrac1{y-2}=\dfrac{x^2}6+\dfrac13[/tex]

    You're asked to find the solution in the form [tex]y=f(x)[/tex], so you should solve for [tex]y[/tex]. You would end up with

    [tex]y=-\dfrac6{x^2+2}+2=\dfrac{2(x^2-1)}{x^2+2}[/tex]

  6. 1.

    [tex]h=-16t^2+vt+c\qquad\text{original equation}\\\\h+16t^2-c=vt\qquad\text{subtract terms not containing v}\\\\v=\dfrac{h+16t^2-c}{t}\qquad\text{divide by the coefficient of v}[/tex]

    2. Filling in the given numbers, we have ...

    [tex]v=\dfrac{131+16(3)^2-5}{3}=90\qquad\text{ft/s}[/tex]

    3. The velocity is undefined when t=0. The equation is telling you it would take infinite velocity to travel some distance in zero time. The equation is meaningless in such a situation.

  7. Initial velocity is 3.5. The equation is s = ut + 1/2at^2, where s - distance, u - inititial velocity, and a - acceleration. How do I modify the acceleration formula to one that gives me initial velocity? You can't change the acceleration formula to one that gives you the initial velocity you want, as a=v/t.

    Step-by-step explanation:

  8. The required formula is [tex]v=\frac{h-c}{t}+16t[/tex]. The initial velocity is 90 ft/s. At t=0, the value of v is undefined.

    Step-by-step explanation:

    The equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v.

    [tex]h=-16t^2+vt+c[/tex]

    We need to solve this equation for v.

    Subtract c from both sides.

    [tex]h-c=-16t^2+vt[/tex]

    Add 16t² on both sides.

    [tex]h-c+16t^2=vt[/tex]

    Divide both sides by t.

    [tex]\frac{h-c+16t^2}{t}=v[/tex]

    [tex]v=\frac{h-c}{t}+16t[/tex]         .... (1)

    Therefore the required formula is [tex]v=\frac{h-c}{t}+16t[/tex].

    We need to find the velocity for

    Initial height c = 5 feet

    Time t = 3 sec

    Height h = 131 feet

    Substitute c = 5, t = 3 and h = 131 in equation (1).

    [tex]v=\frac{131-5}{3}+16(3)[/tex]

    [tex]v=42+48=90[/tex]

    Therefore the initial velocity is 90 ft/s.

    Substitute t=0 in equation (1).

    [tex]v=\frac{h-c}{0}+16(0)[/tex]

    [tex]v=\infty[/tex]

    At t=0, the value of v is undefined.

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