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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Mid-Chapter Quiz

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Exercise 15 Page 132

Substitute h=0 into the function and solve for t.

8.1 sec

The function, h=- t^2+5x+25, tells us the height, h, of the ball after t seconds. To find out how long it takes to reach the bottom, we need to substitute h=0 and solve for t.

h=- t^2+5x+25

h= 0

0 = - t^2+5x+25

Rearrange equation

- t^2+5x+25=0

.LHS /-1.=.RHS /-1.

t^2-5x-25=0

Let's find ( b2)^2 so we can complete the square. (-5/2)^2 = 6.25 Now we can solve by completing the square.

t^2-5x-25=0

▼

Solve for t

LHS+25=RHS+25

t^2 - 5x = 25

LHS+6.25=RHS+6.25

t^2 - 5x + 6.25 = 31.25

Complete the square

(t-2.5)^2 = 31.25

sqrt(LHS)=sqrt(RHS)

t-2.5 ≈ ± 5.59

LHS+2.5=RHS+2.5

t ≈ 2.5 ± 5.59

Since we are measuring time, only the positive answer is relevant. 2.5 + 5.59 = 8.09 It will take the ball about 8.1 seconds to slide down the pole.

Alternative Solution

Solve by Graphing

We can also approximate the answer by looking at where the function's curve crosses the x-axis.

Therefore, like with the algebraic solution, the ball hits the ground 8.1 seconds after it starts.