Question

The height y(in feet) of a ball thrown by a child isy=−114x2+2x+3

where x 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? (Hint: Find y
when x=0)
Your answer is y=____

(b) What is the maximum height of the ball? _____


(c) How far from the child does the ball strike the ground?_____

Answer 1

(a) When the ball is thrown, x = 0. Therefore, we can find the height of the ball when it leaves the child's hand by substituting x = 0 into the equation:

y = -114(0)^2 + 2(0) + 3
y = 3

So the ball is 3 feet high when it leaves the child's hand.

(b) The maximum height of the ball occurs at the vertex of the parabola. We can find the x-coordinate of the vertex using the formula:

x = -b/2a

where a = -114 and b = 2. Substituting these values, we get:

x = -2/(2*(-114)) = 0.0088

We can find the maximum height by substituting this value of x into the equation:

y = -114(0.0088)^2 + 2(0.0088) + 3
y ≈ 3.036

So the maximum height of the ball is approximately 3.036 feet.

(c) To find how far from the child the ball strikes the ground, we need to find the value of x when y = 0. We can solve the equation for x:

-114x^2 + 2x + 3 = 0

Using the quadratic formula, we get:

x = (-2 ± sqrt(2^2 - 4*(-114)*3))/(2*(-114))
x = (-2 ± sqrt(9252))/(-228)

We can ignore the negative root since the ball is moving in the positive x-direction. Simplifying the expression for the positive root, we get:

x = (-2 + sqrt(9252))/(-228) ≈ 0.232

So the ball strikes the ground approximately 0.232 feet away from the child.