Question

Where x is the horizontal distance in feet from the point at which the ball is thrown.How high is the ball when it leaves the childs hand?what is the maximum height of the ball?How far from the child does the ball strike the ground?

Answer 1

the given expression is

`y=(-1)/(14)x^2+4x+3`

here x is the horizontal distance and y is the vertical height.

(a) when the ball just leaves the child's hand the horizontal distance of the ball will be zero

so put x = 0

y = -1/14 (0)^2 + 4(0) + 3

y = 0 + 0 + 3

y = 3

the height of the ball is 3 when it leaves the child's hand.

(b)

the maximum height of the ball

the given equation is the equation of the parabola,

so the max. height of the ball at x = -2b/a

where b is the coefficient of x that is b = 4

a is the coefficient of x^2 that is a = -1/14

so

`\begin{gathered} x=-(2b)/(a) \\ x=(-2*4)/(-(1)/(14)) \\ x=14*8=112 \end{gathered}`

put x= 112 in the equation

`\begin{gathered} y=-(1)/(14)(112)^2+4(112)+3 \\ y=-445 \end{gathered}`

so the maximum height of the ball is 445 ft

(c) for the total horizontal distance of the ball

put y = 0 in the expression because the ball finally touched the ground and at ground y = 0.

so by putting y = 0

you will get a quadratic equation of x

by solving the quadratic equation you will get the value of x