The height \( y \) (in feet) of a ball thrown by a child is \( y=-\frac{1}{16} x^{2}+4 x+5 \) 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? (b) What is the maximum height of the ball? (c) How far from the child does the ball strike the ground?

To answer your questions, let's start by analyzing the quadratic function provided. (a) The height of the ball when it leaves the child's hand is found by evaluating \( y \) when \( x = 0 \). Substituting \( x = 0 \) into the equation, we get: \[ y = -\frac{1}{16}(0)^2 + 4(0) + 5 = 5 \] So, the ball is 5 feet high when it leaves the child's hand. (b) To find the maximum height of the ball, we need to determine the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \), where \( a = -\frac{1}{16} \) and \( b = 4 \): \[ x = -\frac{4}{2(-\frac{1}{16})} = -\frac{4}{-\frac{1}{8}} = 32 \] Now, substituting \( x = 32 \) back into the equation to find the height: \[ y = -\frac{1}{16}(32)^2 + 4(32) + 5 = -\frac{1}{16}(1024) + 128 + 5 = -64 + 128 + 5 = 69 \] Thus, the maximum height of the ball is 69 feet. (c) To find out how far from the child the ball strikes the ground, we set \( y = 0 \) and solve the equation: \[ 0 = -\frac{1}{16}x^2 + 4x + 5 \] Multiplying all terms by -16 to eliminate the fraction gives: \[ 0 = x^2 - 64x - 80 \] Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-(-64) \pm \sqrt{(-64)^2 - 4(1)(-80)}}{2(1)} = \frac{64 \pm \sqrt{4096 + 320}}{2} = \frac{64 \pm \sqrt{4416}}{2} \] Calculating \( \sqrt{4416} \approx 66.4 \): \[ x \approx \frac{64 \pm 66.4}{2} \] We find two possible solutions: \[ x \approx \frac{130.4}{2} \approx 65.2 \quad \text{(the positive root)} \] And \[ x \approx \frac{-2.4}{2} \approx -1.2 \quad \text{(not applicable)} \] Thus, the ball strikes the ground approximately 65.2 feet away from the child.