SOLUTION: A child throws a ball upward from the roof of a house. The tragctory is parabolic, according to the laws of physics. The height the of the ball above the ground, y in meters is giv

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Question 735418: A child throws a ball upward from the roof of a house. The tragctory is parabolic, according to the laws of physics. The height the of the ball above the ground, y in meters is given by the quadratic relation
y=-3x^2+6x+4
a) How high is the roof of the house where the ball is thrown.
b)Complete the square to solve the quadratic equation for the exact values of x when y=0
c) Use the quadratic formula to confirm your answers from part b)
d) Exactly how far away from the house(horizontally) does the ball hit the ground?
e) Use your answers from parts b) and c) and determine the horizontal distance from the house when the ball reaches maximum height.
f) What is the maximum height of the ball?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A child throws a ball upward from the roof of a house.
The trajectory is parabolic, according to the laws of physics.
The height the of the ball above the ground, y in meters is given by the quadratic relation y=-3x^2+6x+4
:
a) How high is the roof of the house where the ball is thrown.
The ball was thrown when x = 0, therefore the roof is 4 meters high
:
b)Complete the square to solve the quadratic equation for the exact values of x when y=0
-3x^2 + 6x + ___ = -4
Coefficient of x has to be 1, divide equation by -3
x^2 - 2x + ___ = 4%2F3
Find the value that completes the square, divide the coefficient of x
and square it, add to both sides
x^2 - 2x + 1 = 4%2F3 + 1
%28x-1%29%5E2 = 7%2F3
Find the square root of both sides
x - 1 = +/-sqrt%287%2F3%29
Two solutions
x = 1 + sqrt%287%2F3%29 ~ 2.53
and
x = 1 - sqrt%287%2F3%29 ~ -.53
:
c) Use the quadratic formula to confirm your answers from part b)
x+=+%28-6+%2B-+sqrt%286%5E2-4%2A-3%2A4+%29%29%2F%282%2A-3%29+
You can do the math on this
:
d) Exactly how far away from the house(horizontally) does the ball hit the ground?
Solution 1 above x ~ 2.53 meters
:
e) Use your answers from parts b) and c) and determine the horizontal distance from the house when the ball reaches maximum height.
That occurs at the axis of symmetry. x = -b/(2a) is the formula for that
x = -6/(2*-3)
x = 1 meter horizontally from the house it will be max height
:
f) What is the maximum height of the ball?
y = -3(1^2) + 6(1) + 4
y = 7 meters max height