The height y (in feet) of a ball thrown by a child is 1/14 x² + 2x + 5

Question

where is the horizontal distance in feet from the point at which the ball is thrown. (a) How high is the ball when it leaves the child's hand? (b) What is the maximum height of the ball? feet feet (c) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01. feet

Answer 1

I am assuming that there should be a negative sign in front of the 1/14 otherwise this problem makes no sense.

(a) The ball is 5 feet above the ground when it leaves the child's hand.

(b) The highest point is found by finding the vertex of the paraboIic path. I will use the -b/2a method.

a = -1/14 b = 2 so -b/2a = -2/[2(-1/14)] = 14.

This is how far the ball has traveled from the child's hand when it reaches it's maximum height.

The maximum height is (-1/14)14^2 + 2(14) + 5 = -14 + 28 + 5 = 19

(c) Now solve the equation for x = 0 which is the height of the ground.

Using the quadratic equation you will get 30.31 feet . (You'll also get a negative answer which is rejected because that mathematically represents the portion of the parabolic path going 'backward' before the ball was thrown.)

Answer 2

general equation is

h(x) = ax^2/2 +bx +c

a=-1/7, b =2, c=5

the constant term c is the intial height = 5 feet

to find max height, take the derivative and set equal to zero

h'(x) = -x/7 +2 = 0

x= -2(-7) =14, plug that into h(x) to get

h(14) = (-1/7)(14^2) +2(14)+5 = 19 feet = max height

set h=0 and solve for x to find distance traveled before hitting the ground

-x^2/14 +2x +5 = 0

multiply all terms by -14 to get

x^2 -28x -70 = 0

complete the square

x^2 -28x +14^2 = 70 +14^2

(x-14)^2 = 266

x = 14 +sqr266

= about 14 + 16.31

= about 30.31 feet when it hits the ground