Answer:
3x²
Step-by-step explanation:
=
=
How are real numbers used to describe real-world situations
1 answer:
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Define
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
