Exercise 14 Page 39

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The height of the rocket is 0 when it returns to the ground.

Let's substitute 0 for h into the given formula.

h=125t-16t^2

Substitute

h= 0

0=125t-16t^2

FactorOut

Factor out t

0=t(125-16t)

To solve this equation, we will apply the Zero Product Property.

0=t(125-16t)

ZeroProdProp

Use the Zero Product Property

lct=0 & (I) 125-16t=0 & (II)

▼

(II): Solve for t

SubEqn

(II): LHS-125=RHS-125

lct=0 & (I) - 16t=- 125 & (II)

DivEqn

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

lct=0 & (I) t= - 125- 16 & (II)

CalcQuot

(II): Calculate quotient

lct=0 & (I) t=7.8125 & (II)

The solutions to the equation are 0 and 7.8125.

In Part B, we have found that t=0 and t=7.8125 are the solutions to the equation. This means that the height of the rocket is zero at t=0 and t=7.8125. Therefore, it will take 7.8125 seconds for rocket to return to the ground.

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Exercises p.39-42

Exercise 14 Page 39

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