Exercise 55 Page 415

We are told that the height h of the object after t seconds of rising and falling with the initial velocity v from the initial height s is given by the following formula. h=- 10t^2+vt+sIt is known that Sherise threw a ball with an initial velocity of 12 meters per second from the height of 54 meters above the ground. Thus, we can substitute v with 12 and s with 54. We want to find in how many seconds the ball will hit the ground and its height above the ground will be 0 meters. Let's substitute 0 for h and solve the equation for t.

h=- 10t^2+vt+s

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Substitute values and simplify

SubstituteValues

Substitute values

0=- 10t^2+ 12t+ 54

RearrangeEqn

Rearrange equation

- 10t^2+12t+54=0

MultEqn

LHS * (- 1)=RHS* (- 1)

10t^2-12t-54=0

We arrived at a quadratic equation. Let's solve it using factoring.

10t^2-12t-54=0

WriteSum

Write as a sum

10t^2-30t+18t-54=0

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Factor out 10t & 18

FactorOut

Factor out 10t

10t(t-3)+18t-54=0

FactorOut

Factor out 18

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10t(t-3)+18(t-3)=0

Factor out (t-3)

(10t+18)(t-3)=0

Use the Zero Product Property

lc10t+18=0 & (I) t-3=0 & (II)

(II): LHS+3=RHS+3

l10t+18=0 t=3

(I): LHS-18=RHS-18

l10t=- 18 t=3

(I): .LHS /10.=.RHS /10.

lt=- 1.8 t=3

Since the time cannot be a negative number, we conclude that the ball will hit the ground after 3 seconds. The answer is A.