Exercise 15 Page 132

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.

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

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Solve for t

AddEqn

LHS+25=RHS+25

t^2 - 5x = 25

AddEqn

LHS+6.25=RHS+6.25

t^2 - 5x + 6.25 = 31.25

CompleteSquare

Complete the square

(t-2.5)^2 = 31.25

SqrtEqn

sqrt(LHS)=sqrt(RHS)

t-2.5 ≈ ± 5.59

AddEqn

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.