Page 760 - Algebra 1

Example
5
Application: Object in Motion
From an initial height s of 70 meters in a stadium, Luis tosses a ball up at an initial velocity v of 5 meters per second. Use the equation -4.9t2 + vt + s = 0 to find the time t when the ball hits the ground.
SOLUTION
Substitute the values into the quadratic formula. Then solve.
-4.9t2 +5t+70=0
2
- b ± √ b - 4 a c
__ t= 2a
= =
=
-(5) ± √(5)2 - 4(-4.9)(70)
___ 2(-4.9)
- 5 ± √ 2 5 + 1 3 7 2 __
-9.8 - 5 ± √ 1 3 9 7
__ -9.8
-5 ± 37.3765 __
≈
t ≈ -3.3037 and t ≈ 4.3241
-9.8 Check
Hint
When the solutions deal with time, we only consider positive values for solutions.
a.
(Ex 1)
b.
(Ex 2)
c.
(Ex 2)
d.
(Ex 3)
e.
(Ex 4)
f.
(Ex 5)
Use the quadratic formula to solve for x.
x2 +3x-18=0
Use the quadratic formula to solve for x. -72-14x+x2 =0
Use the quadratic formula to solve for x.
x2 +80=21x
Use the quadratic formula to solve for x. Then use a graphing calculator to find approximate solutions and verify them. Round the solutions to the nearest ten thousandth.
9x2 +6x-1=0
Use the quadratic formula to solve 4x2 + 5x + 3 = 0 for x.
From an initial height s of 50 meters on a cliff, Janet tosses a ball upward at an initial velocity v of 6 meters/second. At what point does the ball fall back to the ground? Round the solution to nearest ten thousandth.
-4.9(4.3241)2 + 5(4.3241) + 70 ≈ -91.6194 + 21.6205 + 70 ≈ 0 ✓ The ball will land on the ground in approximately 4.3241 seconds.
Lesson Practice
Lesson 110 745