Exercise 49 Page 646

An equation is given to find the height

h

in feet of a projectile

t

seconds after it is shot upwards from ground level.

h = 96 t - 16 t^{2}

We will find the value(s) of

t

when the height of a projectile is

96

feet. Let's first substitute

h = 96

into the equation.

96 = 96 t - 16 t^{2}

Notice that it is a quadratic equation. Let's rewrite the equation so that all of the terms are on the left-hand side and then simplify it as much as possible.

96 = 96 t - 16 t^{2}

AddEqn

$LHS + 16 t^{2} = RHS + 16 t^{2}$

16 t^{2} + 96 = 96 t

SubEqn

$LHS - 96 t = RHS - 96 t$

16 t^{2} - 96 t + 96 = 0

DivEqn

$LHS / 16 = RHS / 16$

t^{2} - 6 t + 6 = 0

Let's recall the Quadratic Formula used to find solution(s) to quadratic equations.

a x^{2} + b x + c = 0 ⇕ x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

To solve our equation we first need to identify the values of

a,

b,

and

c .

t^{2} - 6 t + 6 = 0 ⇕ 1 t^{2} + (- 6) t + 6 = 0

We see that

a = 1,

b = - 6,

and

c = 6 .

Let's substitute these values into the Quadratic Formula.

t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

t = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( 1 ) ( 6 )}{2 ( 1 )}

▼

Simplify

NegNeg

$- (- a) = a$

t = \frac{6 \pm ( - 6 ) ^{2} - 4 ( 1 ) ( 6 )}{2 ( 1 )}

IdPropMult

Identity Property of Multiplication

t = \frac{6 \pm ( - 6 ) ^{2} - 4 ( 6 )}{2}

CalcPow

Calculate power

t = \frac{6 \pm 3 6 - 4 ( 6 )}{2}

Multiply

t = \frac{6 \pm 3 6 - 2 4}{2}

SubTerms

Subtract terms

t = \frac{6 \pm 1 2}{2}

SplitIntoFactors

Split into factors

t = \frac{6 \pm 4 \cdot 3}{2}

SimpRoot

Simplify root

t = \frac{6 \pm 2 3}{2}

FactorOut

Factor out $2$

t = \frac{2 ( 3 \pm 3 )}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

t = 3 \pm 3

The solutions for this equation are

t = 3 \pm 3 .

Let's calculate them separately using a calculator and round to the nearest tenth.

t_{1} = 3 + 3 t_{2} = 3 - 3

CalcRoot

Calculate root

t_{1} = 3 + 1.732050 \dots t_{2} = 3 - 1.732050 \dots

AddSubTerms

Add and subtract terms

t_{1} = 4.732050 \dots t_{2} = 1.267949 \dots

RoundDec

Round to $1$ decimal place(s)

t_{1} = 4.7 t_{2} = 1.3

Therefore, the height of the projectile will be

96

feet on the way up at about

1.3

seconds and on the way down at about

4.7

seconds.

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