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2 years ago

20 POINTS

Mathematics

2 answers:

Anna35 [415]2 years ago

8 0

Answer:

Step-by-step explanation:

Our function, in terms of the gravitational pull in the metric system, will be

$s(t)=-4.9t^2+12t+2$

where the leading coefficient represents the acceleration due to gravity (if you do not know calculus, just take me at my word on this!), 12 is the initial vertical velocity, and 2 is the initial height of the baseball. Being asked the question, "How long will it take for the ball to hit the ground?" is the exact same as being asked, "How long was the ball in the air?" Be aware of the wording!!

s(t) represents the height of the ball at a certain time during its flight. If the ball hits the ground, we know that the height of something ON the ground is 0, so sub in 0 for s(t) and factor to solve for t.

If you throw this into the quadratic formula, you get that

t = -.2 seconds and t = 2.6 seconds.

Since time will NEVER be negative, we know that it takes the ball 2.6 seconds to hit the ground. It was also, then, in the air for 2.6 seconds.

Montano1993 [528]2 years ago

4 0

Answer:

2.6 seconds

Step-by-step explanation:

g = 10

s = 2

u = -12

t = ?

2 = -12t + 5t²

5t² - 12t - 2 = 0

t = [12 +/- sqrt(12² - 4(5)(-2)]/2(5)

t = (12 +/- sqrt(184))/10

t = 2.55646599663

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3 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

2 years ago