Exercise 46 Page 40

Practice makes perfect

Let's factor the right hand side of the given formula.

263t-16t^2

SplitIntoFactors

Split into factors

263* t- 16t * t

FactorOut

Factor out the GCF

t(263-16t)

This expression represents the height in factored form.

In Part A, we have written the formula in factored form.

h=263t-16t^2 ⇒ h=t(263-16t)

We will first substitute 0 for h. Then, to solve the equation, we will apply the Zero Product Property.

h=t(263-16t)

Substitute

h= 0

0=t(263-16t)

ZeroProdProp

Use the Zero Product Property

lct=0 & (I) 263-16t=0 & (II)

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Solve for t

SubEqn

(II): LHS-263=RHS-263

lct=0 & (I) - 16t=- 263 & (II)

DivEqn

(II): .LHS /(- 16).=.RHS /(- 16).

lct=0 & (I) t=- 263/- 16 & (II)

CalcQuot

(II): Calculate quotient

lct=0 & (I) t=16.4375 & (II)

The solutions to the equation are 0 and 16.4375. t=0 means that the shell is not fired, and therefore, the height of is zero. t=16,4375 means that 16.4375 seconds after the shell is fired, it returns to the ground. As a result, the answers are practical.

To find the height of the shell after 8 second, we need to substitute 8 for t in the equation.

h=263t-16t^2

Substitute

t= 8

h=263( 8)-16( 8)^2

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Simplify right-hand side

CalcPow

Calculate power

h=263(8)-16(64)

Multiply

h=2104-1024

SubTerm

Subtract term

h=1080

The height of the shell is 1080ft after 8 second. Let's find its height after 10 seconds.

h=263t-16t^2

Substitute

t= 10

h=263( 10)-16( 10)^2

▼

Simplify right-hand side

CalcPow

Calculate power

h=263(10)-16(100)

Multiply

h=2630-1600

SubTerm

Subtract term

h=1030

The height of the shell is 1030ft after 10 seconds.

We have found that the shell has a height of 1080 ft after 8 seconds. After two seconds, its height decreases to 1030 ft. Therefore, the shell is falling.

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