The height y (in feet) of a ball thrown by a child is y=-1/14x^2 + 4x + 5 ...?

Question

The height y (in feet) of a ball thrown by a child is y = -1/14x^2 + 4x + 5 where x is the horizonal distance in feet from the point at which the ball is thrown.
1.) What is the maximum height of the ball?
2.) How far from the child does the ball strike the ground?

Answer 1

maximum height occurs at the vertex of the parabola since it opens downward.
Vertex: x = -b/2a = -(4)/(2(-1/14)) = 4/(1/7) = 4(7) = 28
y = -1/14(28)^2 + 4(28) + 5 = 61
max height is 61 ft

Ball strikes ground when height = 0, so put a 0 in for y and solve
0 = -1/14x^2 + 4x + 5
multiply by -14 to clear fraction
0 = x^2 - 56x - 70
x = -1.22 or 57.22 using the quadratic formula.

57.2 ft from child

Answer 2

Hello,
take the derivative giving us

y' = -1/7x +4 =0 so x = 28

Now put 28 in for x to get the max height -56 + 96 +5 = 157 Feet

Then set the equation = 0 to find the point at which it hits the ground. 0 = -1/14x^2 +4x +5 and multiply by 14 to get rid of the denominator 0 = -x^2 + 56x + 70 or x^2 -56x -70 = 0 so
Use the quadratic formula to find x.

Hope This Helps!

Answer 3

y = -1/14x^2 + 4x + 5 . . . . differentiating
y ' = - 1 /7 x + 4 = 0
x / 7 = 4
x = 28 . . . horizontal distance

maximum height
y = -1/14(28)^2 + 4(28) + 5
y = 173 feet . . . maximum height

distance when it strike the ground
y = -1/14x^2 + 4x + 5 = 0
x^2 - 56 x - 70 = 0
solving x = 57.223 feet

Answer 4

y = -1/14x^2 + 4x + 5
y (') = - 1 /7 x + 4 = 0
x / 7 = 4
x = 28

Maximum height:
y = -1/14(28)^2 + 4(28) + 5
y = 173 feet which is maximum height

Distance when it strike the ground
y = -1/14x^2 + 4x + 5 = 0
x^2 - 56 x - 70 = 0
solving x = 57.223 feet

I HOPE THIS HELPS