205x2 + 32 te

12. The path of a baseball thrown at an angle of 40° can be modeled by y = -0.05x + 3.2% to

the horizontal distance in feet from the release point and v is the corresponding height, in feet. Calculate the

average rate of change of the height over the interval 8 < X < 20.

the

(1) 3

(2)1/3

(3)5/9

(4)9/5

Question

23-01-2020
Mathematics

contestada

205x2 + 32 te

12. The path of a baseball thrown at an angle of 40° can be modeled by y = -0.05x + 3.2% to

the horizontal distance in feet from the release point and v is the corresponding height, in feet. Calculate the

average rate of change of the height over the interval 8 < X < 20.

the

(1) 3

(2)1/3

(3)5/9

(4)9/5

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ElderVujadin ElderVujadin · Answer 1 · 2020-01-23T16:11:03+01:00

Answer: If the equation of motion is of the form of [tex]y=c+ax-bx^2[/tex], then the average height is [tex]h_{ave}=c+\frac{1}{2}a(x_1+x_2)-\frac{1}{3}b(x_1^2+x_1x_2+x_2^2).[/tex]

Step-by-step explanation: If the baseball is thrown from the initial height of [tex]h_0[/tex] with the speed of [tex]v[/tex] at an angle of [tex]40^\circ[/tex], then the equation of motion that connects the height [tex]y[/tex] and the horizontal distance [tex]x[/tex] is [tex]y=h_0+x\tan40^\circ-\frac{g}{2v^2cos^240^\circ}x^2[/tex], where [tex]g=9.8\text{ m/s}^2[/tex] is the gravitational acceleration. For simplicity, we will take that [tex]y=c+ax-bx^2[/tex], where [tex]a,b,c[/tex] are some constants. The average height between [tex]x=x_1[/tex] (here [tex]x_1=8[/tex]) and [tex]x=x_2[/tex] (here [tex]x_2=20[/tex]) is given by the formula

[tex]h_{ave}=\frac{1}{x_2-x_1}\int_{x_1}^{x_2}y(x)dx=\frac{1}{x_2-x_1}\int_{x_1}^{x_2}(c+ax-bx^2)dx.[/tex]

This further yields, after integration,

[tex]h_{ave}=\frac{1}{x_2-x_1}\left(cx+a\frac{x^2}{a}-b\frac{x^3}{2}\right)\limits_{x_1}^{x_2}=c+\frac{1}{2}a(x_1+x_2)-\frac{1}{3}b(x_1^2+x_1x_2+x_2^2).[/tex]