Question 1021181
he height y (in feet) of a ball thrown by a child is
y={{{-1/16}}}{{{x^2+2x+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? feet
When the ball leaves his hand, horizontal distance is 0, therefore x=0
If you substitute 0 for x in the above equation, y = 5, ball is 5' high then
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(b) What is the maximum height of the ball? feet
Maximum height occurs on the axis of symmetry of the equation
We can find the axis of symmetry using x = -b/(2a)
In this equation a=-1/16, b=2
x = {{{(-2)/(2*(-1/16))}}} = {{{(-2)/(-1/8)}}}
invert the dividing fraction and multiply and you have
x = 16 ft is the horizontal distance for max height
Find the maximum height by substituting 16 for x in the given equation
y={{{-1/16}}}{{{16^2+2(16)+5}}}
y = -16 + 32 + 5
y = 21 ft is the max height
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(c) How far from the child does the ball strike the ground? feet
When the ball hits the ground y = 0, therefore
{{{-1/16}}}{{{x^2+2x+5}}} = 0
Use the quadratic formula to find x,
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
1/16 to -.0625, easier to deal with, a=-.0625; b=2; c=5
{{{x = (-2 +- sqrt( 2^2-4*-.0625*5 ))/(2*-.0625) }}}
I'll let you do the math here
I got the positive solution of x = 34.33 ft from the child to where it hits the ground
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looks like this
{{{ graph( 300, 200, -6, 40, -10, 30, -.0625x^2+2x+5) }}}
note that when x=0, y=5, (5ft off the ground when the child throws the ball)
When x=16 (16 ft from the child) the ball reaches max height (y=21')
When x about 34ft from the child, the ball strikes the ground, (y=0)
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Was this understandable to you. CK, ankor@att.net