Question

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

Answer 1

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

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

`h_(ave)=(1)/(x_2-x_1)\int_(x_1)^(x_2)y(x)dx=(1)/(x_2-x_1)\int_(x_1)^(x_2)(c+ax-bx^2)dx.`

This further yields, after integration,

`h_(ave)=(1)/(x_2-x_1)\left(cx+a(x^2)/(a)-b(x^3)/(2)\right)\limits_(x_1)^(x_2)=c+(1)/(2)a(x_1+x_2)-(1)/(3)b(x_1^2+x_1x_2+x_2^2).`