Exercise 11 Page 68

We can begin by identifying that h_0 is the initial height from which we are dropping the brush. When the brush hits the ground is h. Therefore, we can substitute 0 for h and 6 for h_0. h=-16t^2+ h_0 ⇒ 0=-16t^2+ 6 Next, we can solve for t.

0=-16t^2+6

SubEqn

LHS-6=RHS-6

-6=-16t^2

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Solve for t

DivEqn

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

6/16=t^2

ReduceFrac

a/b=.a /2./.b /2.

3/8=t^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

±sqrt(3/8)=t

CalcRoot

Calculate root

± 0.6≈ t

RearrangeEqn

Rearrange equation

t≈± 0.6

Mathematically our solution can be positive or negative. However, a negative solution does not make sense in this scenario. Our solution must then be t=0.6.