McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
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Exercise 70 Page 176

Practice makes perfect
a When a ball is kicked, it follows a parabolic route. It first rises for a certain period of time until it reaches the maximum height and then it begins to fall. Let's model this situation.

The x-coordinate of the vertex will give us the rise time. We can find it by using the formula for the axis of symmetry, t= - b2a. Let's rewrite the function in the general form of a quadratic function to determine a and b. f(x)= ax^2+ bx+ c ⇓ h(t)= -4.9t^2+ 14.7t+ 0 Now, we can find the rise time.

t= -b/2a
t= -14.7/2( -4.9)
t= -14.7/-9.8
t= 14.7/9.8
t= 3/2

The rise time is 32 seconds. Therefore, the fall time will be also 32 seconds.

Flight time: 3/2+3/2=3 Therefore, the ball will in the air for 3 seconds.

b In Part A, we found that the ball will be in the air for 3 seconds. By substituting t=3 in the distance function, we can find the distance traveled.

d(t)=16t
d( 3)=16( 3)
d(3)=48

The ball travels 48 meters before it hits the ground.

c The maximum height of the ball will be on the vertex of its parabolic route. We found the x-coordinate of the vertex in Part A as t= 32. By substituting this value into the height function, we can find the maximum height.

h(t)=-4.9t^2+14.7t
h( 3/2)=-4.9( 3/2)^2+14.7( 3/2)
h(3/2)=-4.9(9/4)+14.7(3/2)
h(3/2)=-11.025+22.05
h(3/2)=11.025

Therefore, the maximum height is 11.025 meters.