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McGraw Hill Glencoe Algebra 1, 2012 View details

Mid-Chapter Quiz

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Exercise 8 Page 582

a We are given the equation

h = - 16 t^{2} + 90 t,

with

h

as the balls height after

t seconds .

For this part, we need to know the height of the ball after one second. Let's substitute and find

h .

h = - 16 t^{2} + 90 t

Substitute

$t = 1$

h = - 16 (1)^{2} + 90 (1)

CalcPowProd

Calculate power and product

h = - 16 + 90

AddTerms

Add terms

h = 74

After one second, the ball will be

74 feet

off the ground.

b Since this is a quadratic equation, we can use the vertex of the parabola as a maximum point. The first step in finding the vertex is to find the axis of symmetry for the function

h = - 16 t^{2} + 90 t

t = - \frac{b}{2 a}

SubstituteII

$a = - 16$ , $b = 90$

t = - \frac{9 0}{2 ( - 1 6 )}

▼

Simplify right-hand side

Multiply

t = - \frac{9 0}{- 3 2}

CalcQuot

Calculate quotient

t = - (- 2.8125)

NegNeg

$- (- a) = a$

t = 2.8125

Since the question asks for time to get to the maximum, we do not need the other coordinate of the vertex. It takes the ball about

2.8 seconds

to reach it's maximum height.

c The most common way to find the amount of time it takes to reach

0 ft

is the let

h = 0

and solve for

t .

h = - 16 t^{2} + 90 t

Substitute

$h = 0$

0 = - 16 t^{2} + 90 t

▼

Solve for

t

FactorOut

Factor out $- 2 t$

0 = - 2 t (8 t - 45)

ZeroProdProp

Use the Zero Product Property

- 2 t = 0 8 t - 45 = 0 (I) (II)

DivEqn

$(I):$ $LHS / - 2 = RHS / - 2$

t = 0 8 t - 45 = 0

AddEqn

$(II):$ $LHS + 45 = RHS + 45$

t = 0 8 t = 45

DivEqn

$(II):$ $LHS / 8 = RHS / 8$

t = 0 t = 5.625

The ball is

0 ft

off the ground at

0 sec

and at about

5.6 sec .

These represent the times the ball is on the ground, first when the player kicks the ball, then when the ball lands.

Alternative Solution

Using Reasoning

In Part B, we determined that the ball takes about

2.8 sec

to reach the vertex. Since the parabola is symmetric about the axis of symmetry, we know that it will take the same amount of time again to get back to the same height.

2.8 \times 2 = 5.6

The player kicked it from ground level,

0 ft

off the ground, so we know it will reach ground level again after

5.6 seconds .