the height y (in feet) of a ball thrown by a child is y=−1/14x2+4x+5

Question

The height y (in feet) of a ball thrown by a child is y=−1/14x2+4x+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

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

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

Answer 1

(a)

at time zero, t=0, the distance is 5 feet, so the boy is 5 feet tall.

(-1/14) x^2 + 4x + 5 =0

x^2 - 56x - 70 = 0 <--- multiplies by -14

x= [ 56 +or- sqrt( 3136 + 280)] / 2

= [ 56 +or- sqrt( 3416)]/2

(b)

the max occurs at the average of the solutions, which is 28.

or -b/2a = -4/(2*-1/14) = -4/ (-1/7) = 28

(1/14)*28^2 +4*28 + 5 = 61

so the max height is 61 feet after 28 feet

(c)

the ball will hit the ground at (56 + sqrt(3416)) / 2 = 57.223