The height y in feet of a ball thrown by a child is y 112 x

The height y (in feet) of a ball thrown by a child is y = -1/12 x^3 + 2x + 3 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 child\'s hand? Your answer is y = What is the maximum height of the ball? How far from the child does the ball strike the ground?

Solution

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Given

y = (-1/12)x^2 + 2x + 3

a) Now when ball is relese from hand, i.e. when x = 0

y = 3

Therefore height of ball when it is relesed from hand is 3 feet.

b) In order to calculate maximum height of ball, let us diferentiate y w.r.t. x

dy/dx = (-1/6)x + 2 = 0

=> x =12 feet

Nw if x = 12 corresponds to horizntal distance w.r.t. maximum height attained by ball, then o double differeentiation the result should be negative.

d2y/dx2 = (-1/6)

Hence calculating y corresponding to x = 12

y = (-1/12)x^2 + 2x + 3

= ((-1/12)12^2 + 2*12 + 3

= -12 + 24 + 3

= 15 feet

Maximum height = 15 feet

c) Now when ball strikes the ground, i.e. when y = 0

(-1/12)x^2 + 2x + 3 = 0

-(x^2) + 24x + 36 = 0

x^2 -24x - 36 = 0

x = [24 + sqrt (576 + 4*36)] / 2

x = 25.4 feet

Solution

Solution